Higher-Order Linear Differential Equations

This chapter focuses on the systematic methodologies for solving higher-order linear differential equations, a cornerstone topic in advanced calculus. It thoroughly examines equations with constant coefficients, the Cauchy-Euler equation, and the method of variation of parameters. Proficiency in these analytical techniques is essential for the CUET PG MA examination, as they represent frequently tested concepts within the Differential Equations component.

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Chapter Contents

| # | Topic |
|---|-------|
| 1 | Equations with Constant Coefficients |
| 2 | Cauchy-Euler Equation |
| 3 | Method of Variation of Parameters |

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We begin with Equations with Constant Coefficients.

Part 1: Equations with Constant Coefficients

We investigate linear ordinary differential equations where the coefficients of the derivatives are constants. These equations are fundamental in various scientific and engineering disciplines, providing models for systems ranging from electrical circuits to population dynamics. Mastery of their solution techniques is essential for the CUET PG examination.

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Core Concepts

1. General Form and Classification of ODEs

A general $n$-th order ordinary differential equation can be expressed as

We classify ODEs based on their order, linearity, and homogeneity.

Quick Example: Classify the ODE

Step 1: Determine the order.
> The highest derivative is $\frac{d^2y}{dx^2}$, so the order is 2.

Step 2: Check for linearity.
> The term $y^2$ makes the equation non-linear.
>
> The equation is a second-order non-linear homogeneous ODE.

:::question type="MCQ" question="Classify the differential equation

" options=["Third-order linear non-homogeneous","Third-order non-linear homogeneous","Third-order non-linear non-homogeneous","Second-order linear non-homogeneous"] answer="Third-order non-linear non-homogeneous" hint="Check the highest derivative for order, and powers/functions of $y$ or its derivatives for linearity. The $e^x$ term indicates non-homogeneity." solution="Step 1: Order
The highest derivative is $\frac{d^3y}{dx^3}$, so the order is 3.

Step 2: Linearity
The term $\sin(y)$ is a non-linear function of $y$. Therefore, the equation is non-linear.

Step 3: Homogeneity
The term $e^x$ on the right-hand side is a non-zero function of $x$. Therefore, the equation is non-homogeneous.

Combining these, the equation is a third-order non-linear non-homogeneous differential equation."
:::

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2. Homogeneous Linear ODEs with Constant Coefficients

We consider equations of the form

where $a_i$ are constants. The general solution, known as the Complementary Function (CF), is obtained by solving the characteristic equation.

The dimension of the solution space for an $n$-th order linear homogeneous ODE is $n$. This means its general solution will be a linear combination of $n$ linearly independent solutions.

Quick Example: Determine the dimension of the solution space for

Step 1: Identify the order of the ODE.
> The highest derivative is $y'''$, so the order is 3.

Step 2: Apply the dimension rule.
> For an $n$-th order linear homogeneous ODE, the dimension of the solution space is $n$.
>
> Thus, the dimension of the solution space is 3.

:::question type="MCQ" question="The dimension of the solution space for the differential equation

is:" options=["1","2","3","4"] answer="4" hint="The dimension of the solution space for an $n$-th order linear homogeneous ODE is $n$." solution="The given differential equation is

This is a fourth-order linear homogeneous differential equation.
The dimension of the solution space for an $n$-th order linear homogeneous ODE is $n$.
Here, $n=4$.
Therefore, the dimension of the solution space is 4."
:::

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3. Cases for Roots of the Characteristic Equation

The form of the complementary function $y_c(x)$ depends critically on the nature of the roots of the characteristic equation.

3.1. Case 1: Distinct Real Roots

If the characteristic equation has $n$ distinct real roots $m_1$, $m_2$, \dots, $m_n$, then the $n$ linearly independent solutions are $e^{m_1x}$, $e^{m_2x}$, \dots, $e^{m_nx}$.

Quick Example: Find the general solution of

Step 1: Form the characteristic equation.

Step 2: Solve for the roots.

Step 3: Write the general solution.

:::question type="MCQ" question="A complete solution of

is

The respective values of $a_1$ and $a_2$ are:" options=["$-7, 10$","$7, -10$","$-7, -10$","$7, 10$"] answer="$-7, 10$" hint="If the roots are $m_1$ and $m_2$, the characteristic equation is $(m-m_1)(m-m_2)=0$. Expand this to find $a_1$ and $a_2$." solution="Step 1: Identify the roots from the given solution.
The given solution is

This implies the roots of the characteristic equation are $m_1 = 2$ and $m_2 = 5$.

Step 2: Form the characteristic equation.
The characteristic equation is $(m - m_1)(m - m_2) = 0$.

Step 3: Compare with the general form.
The characteristic equation for

is

Comparing

with

we find:
$a_1 = -7$
$a_2 = 10$

Thus, the respective values of $a_1$ and $a_2$ are $-7$ and $10$."
:::

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3.2. Case 2: Repeated Real Roots

If a real root $m$ has multiplicity $k$, then the $k$ linearly independent solutions associated with this root are $e^{mx}, xe^{mx}, x^2e^{mx}, \dots, x^{k-1}e^{mx}$.

Quick Example: Find the general solution of $y'' + 4y' + 4y = 0$.

Step 1: Form the characteristic equation.
>

Step 2: Solve for the roots.
>

>

Step 3: Write the general solution.
>

:::question type="MCQ" question="The general solution of the differential equation:

is:" options=["$c_1e^x + c_2e^{-x} + c_3xe^{-x}$","$c_1e^x + c_2xe^x + c_3x^2e^x$","$c_1e^{-x} + c_2xe^{-x} + c_3x^2e^{-x}$","$c_1e^x + c_2x^2e^x + c_3x^3e^x$"] answer="$c_1e^x + c_2xe^x + c_3x^2e^x$" hint="Recognize the characteristic polynomial as a binomial expansion. The roots will be repeated." solution="Step 1: Form the characteristic equation.
The characteristic equation is:

Step 2: Factor the characteristic equation.
This is a standard binomial expansion:

So,

Step 3: Identify the roots and their multiplicity.
The root is $m = 1$ with multiplicity 3.

Step 4: Write the general solution.
For a root $m$ with multiplicity $k$, the solutions are $e^{mx}, xe^{mx}, \dots, x^{k-1}e^{mx}$.
Here, $m=1$ and $k=3$. So the solutions are $e^x, xe^x, x^2e^x$.
The general solution is:

Answer: $\boxed{c_1e^x + c_2xe^x + c_3x^2e^x}$"
:::

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3.3. Case 3: Complex Conjugate Roots

If the characteristic equation has complex conjugate roots of the form $m = \alpha \pm i\beta$, then the two linearly independent solutions are $e^{\alpha x}\cos(\beta x)$ and $e^{\alpha x}\sin(\beta x)$.

Quick Example: Find the general solution of $y'' + 2y' + 5y = 0$.

Step 1: Form the characteristic equation.
>

Step 2: Solve for the roots using the quadratic formula:
>

>

Step 3: Identify $\alpha$ and $\beta$.
> Here, $\alpha = -1$ and $\beta = 2$.

Step 4: Write the general solution.
>

:::question type="MCQ" question="The complete solution of the differential equation:

is:" options=["$y = c_1 e^x + c_2 \cos x + c_3 \sin x$","$y = c_1 e^{-x} + c_2 \cos x + c_3 \sin x$","$y = c_1 e^{-x} + c_2 \cos 2x + c_3 \sin 2x$","$y = c_1 e^x + c_2 \cos 2x + c_3 \sin 2x$"] answer="$y = c_1 e^{-x} + c_2 \cos 2x + c_3 \sin 2x$" hint="Factor the characteristic polynomial to find one real root and a pair of complex conjugate roots." solution="Step 1: Form the characteristic equation.

Step 2: Factor the characteristic equation.
We can factor by grouping:

Step 3: Find the roots.
From $(m+1) = 0$, we get $m_1 = -1$.
From $(m^2+4) = 0$, we get $m^2 = -4$, so $m = \pm \sqrt{-4} = \pm 2i$.
So, the roots are $m_1 = -1$, $m_2 = 2i$, $m_3 = -2i$.

Step 4: Write the general solution.
For the real root $m_1 = -1$, the solution is $c_1 e^{-x}$.
For the complex conjugate roots $m = \pm 2i$, we have $\alpha = 0$ and $\beta = 2$. The solution is $e^{0x}(c_2 \cos(2x) + c_3 \sin(2x)) = c_2 \cos(2x) + c_3 \sin(2x)$.
Combining these, the complete solution is:

Answer: $\boxed{y = c_1 e^{-x} + c_2 \cos(2x) + c_3 \sin(2x)}$"
:::

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4. Special Solution Properties

Certain homogeneous ODEs, particularly $y''+y=0$, lead to solutions with useful properties. For $y''+y=0$, the solution is $y(x) = c_1 \cos x + c_2 \sin x$. If $y(x)=f(x)$, then $y'(x)=f'(x)=g(x)$. We observe that $f''(x) = -f(x)$, and $g'(x) = f''(x) = -f(x)$.

Consider the function $h(x) = [f(x)]^2 + [g(x)]^2$.
Differentiating $h(x)$ with respect to $x$:

Since $h'(x)=0$, $h(x)$ must be a constant. This property can be used to solve specific problems.

Quick Example: If $f''(x) = -f(x)$, $f'(x) = g(x)$, $h(x) = [f(x)]^2 + [g(x)]^2$ and $h(0) = 5$, find $h(\pi)$.

Step 1: Establish the relationship between $f(x)$ and $g(x)$.
> Given $f''(x) = -f(x)$ and $f'(x) = g(x)$. This implies $g'(x) = f''(x) = -f(x)$.

Step 2: Differentiate $h(x)$.
>

Step 3: Substitute the relationships.
>

Step 4: Conclude the value of $h(x)$.
> Since $h'(x) = 0$, $h(x)$ is a constant.
> Given $h(0) = 5$, then $h(\pi)$ must also be 5.

:::question type="MCQ" question="Let $y(x)$ be a solution to the differential equation $y'' + 9y = 0$. If $z(x) = [y(x)]^2 + \frac{1}{9}[y'(x)]^2$ and $z(0) = 4$, then $z(\pi/2)$ is:" options=["$1/9$","$4/9$","$4$","$9$"] answer="$4$" hint="Differentiate $z(x)$ with respect to $x$ and use the given differential equation to simplify. If $z'(x)=0$, then $z(x)$ is a constant." solution="Step 1: Use the given differential equation to find $y''(x)$.
From $y'' + 9y = 0$, we have $y'' = -9y$.

Step 2: Differentiate $z(x)$ with respect to $x$.

Step 3: Substitute $y''(x) = -9y(x)$ into $z'(x)$.

Step 4: Conclude the value of $z(x)$.
Since $z'(x) = 0$, $z(x)$ is a constant function.
Given $z(0) = 4$, it follows that $z(x) = 4$ for all $x$.
Therefore, $z(\pi/2) = 4$.
Answer: $\boxed{4}$"
:::

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Non-Homogeneous Linear ODEs with Constant Coefficients

For a non-homogeneous linear ODE $a_n y^{(n)} + \dots + a_0 y = F(x)$, the general solution is the sum of the complementary function ($y_c$) and a particular integral ($y_p$).

We have already discussed methods for finding $y_c$. We now focus on determining $y_p$.

1. Method of Undetermined Coefficients

This method is applicable when $F(x)$ is a polynomial, exponential function, sine or cosine function, or finite sums/products of these. The form of $y_p$ is assumed based on $F(x)$ and then coefficients are determined by substitution into the ODE.

We summarize the forms of $y_p$ for common $F(x)$:

| $F(x)$ | Form of $y_p$ (Initial Guess) | Modification for Resonance ($y_p = x^s \cdot (\text{Initial Guess})$) |
| :---------------------------------------- | :------------------------------------------------ | :----------------------------------------------------------------------------------------- |
| $P_n(x) = A_n x^n + \dots + A_0$ | $Q_n(x) = B_n x^n + \dots + B_0$ | $s=0$ if $m=0$ is not a root of CF, $s=k$ if $m=0$ is a root of CF with multiplicity $k$. |
| $Ae^{\alpha x}$ | $Be^{\alpha x}$ | $s=0$ if $\alpha$ is not a root of CF, $s=k$ if $\alpha$ is a root of CF with multiplicity $k$. |
| $A\sin(\beta x)$ or $A\cos(\beta x)$ | $B\cos(\beta x) + C\sin(\beta x)$ | $s=0$ if $\pm i\beta$ are not roots of CF, $s=k$ if $\pm i\beta$ are roots of CF with multiplicity $k$. |
| $P_n(x)e^{\alpha x}$ | $Q_n(x)e^{\alpha x}$ | $s=0$ if $\alpha$ is not a root of CF, $s=k$ if $\alpha$ is a root of CF with multiplicity $k$. |
| $P_n(x)\sin(\beta x)$ or $P_n(x)\cos(\beta x)$ | $(Q_n(x)\cos(\beta x) + R_n(x)\sin(\beta x))$ | $s=0$ if $\pm i\beta$ are not roots of CF, $s=k$ if $\pm i\beta$ are roots of CF with multiplicity $k$. |
| $e^{\alpha x}\sin(\beta x)$ or $e^{\alpha x}\cos(\beta x)$ | $e^{\alpha x}(B\cos(\beta x) + C\sin(\beta x))$ | $s=0$ if $\alpha \pm i\beta$ are not roots of CF, $s=k$ if $\alpha \pm i\beta$ are roots of CF with multiplicity $k$. |

Note: If $F(x)$ is a sum of terms, $y_p$ is the sum of particular integrals for each term.

1.1. $F(x) = Ae^{\alpha x}$

Quick Example: Find the PI of $y'' - 4y = 3e^{2x}$.

Step 1: Find the roots of the characteristic equation (for CF).
>

> The roots are $m_1 = 2, m_2 = -2$.

Step 2: Determine the initial form of $y_p$.
> $F(x) = 3e^{2x}$. The value $\alpha = 2$.
> Since $\alpha = 2$ is a root of the characteristic equation with multiplicity $s=1$, we must multiply by $x$.
> Initial guess: $y_p = Ae^{2x}$.
> Modified guess: $y_p = Axe^{2x}$.

Step 3: Find derivatives of $y_p$.
>

Step 4: Substitute $y_p, y_p'$ into the ODE.
>

Step 5: Equate coefficients to find $A$.
>

Step 6: Write the particular integral.
>

:::question type="MCQ" question="The particular integral (PI) of the differential equation:

is given as:" options=["$xe^{(3/2)x}$","$e^{(3/2)x}$","$xe^{-(3/2)x}$","$e^{-(3/2)x}$"] answer="$xe^{(3/2)x}$" hint="First, find the roots of the characteristic equation. Check if the exponent in $F(x)$ is one of the roots (resonance case)." solution="Step 1: Find the roots of the characteristic equation.
The characteristic equation is:

Using the quadratic formula:


The roots are $m_1 = \frac{1+5}{4} = \frac{6}{4} = \frac{3}{2}$ and $m_2 = \frac{1-5}{4} = \frac{-4}{4} = -1$.

Step 2: Determine the form of $F(x)$ and check for resonance.
The forcing function is $F(x) = 5e^{(3/2)x}$.
The exponent is $\alpha = 3/2$.
Since $\alpha = 3/2$ is a root of the characteristic equation (specifically, $m_1 = 3/2$) with multiplicity 1, we must multiply the initial guess by $x$.

Step 3: Propose the particular integral $y_p$.
Initial guess: $y_p = Ae^{(3/2)x}$.
Modified guess (due to resonance): $y_p = Axe^{(3/2)x}$.

Step 4: Calculate derivatives of $y_p$.

Step 5: Substitute $y_p, y_p'$ and $y_p''$ into the differential equation.

Step 6: Solve for $A$.

Step 7: Write the particular integral.

Answer: $\boxed{xe^{(3/2)x}}$"
:::

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1.2. $F(x) = A\sin(\beta x)$ or $A\cos(\beta x)$

Quick Example: Find the PI of $y'' + y = \sin x$.

Step 1: Find the roots of the characteristic equation.
>

> The roots are $m_1 = i, m_2 = -i$.

Step 2: Determine the initial form of $y_p$.
> $F(x) = \sin x$. Here, $\beta = 1$. The complex roots of the characteristic equation are $\pm i\beta = \pm i(1)$.
> Since $\pm i$ are roots of the characteristic equation with multiplicity $s=1$, we must multiply by $x$.
> Initial guess: $y_p = A\cos x + B\sin x$.
> Modified guess: $y_p = x(A\cos x + B\sin x)$.

Step 3: Find derivatives of $y_p$.
>

Step 4: Substitute $y_p, y_p''$ into the ODE.
>

Step 5: Equate coefficients.
> Coefficients of $\sin x$: $-2A = 1 \Rightarrow A = -\frac{1}{2}$
> Coefficients of $\cos x$: $2B = 0 \Rightarrow B = 0$

Step 6: Write the particular integral.
>

:::question type="MCQ" question="For the differential equation:

the particular integral is:" options=["$-\frac{5}{17}\sin x + \frac{3}{17}\cos x$","$\frac{2}{13}\sin x - \frac{10}{13}\cos x$","$\frac{1}{5}\sin x + \frac{3}{5}\cos x$","$\frac{5}{17}\sin x - \frac{3}{17}\cos x$"] answer="$-\frac{5}{17}\sin x + \frac{3}{17}\cos x$" hint="Find the roots of the characteristic equation. Since $i$ is not a root, no modification is needed for the PI. Substitute $y_p = A\cos x + B\sin x$." solution="Step 1: Find the roots of the characteristic equation.
The characteristic equation is:


The roots are $m_1 = 4, m_2 = -1$.

Step 2: Determine the form of $F(x)$ and check for resonance.
The forcing function is $F(x) = 2\sin x$. Here, $\beta = 1$.
The roots $\pm i\beta = \pm i$ are not roots of the characteristic equation. So, no modification is needed.

Step 3: Propose the particular integral $y_p$.

Step 4: Calculate derivatives of $y_p$.

Step 5: Substitute $y_p, y_p', y_p''$ into the differential equation.

Step 6: Equate coefficients.
Comparing coefficients of $\sin x$:

Comparing coefficients of $\cos x$:

From (2), $3B = -5A \Rightarrow B = -\frac{5}{3}A$.

Substitute $B$ into (1):

Now find $B$:

Step 7: Write the particular integral.

Answer: $\boxed{-\frac{5}{17}\sin x + \frac{3}{17}\cos x}$"
:::

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1.3. $F(x) = P_n(x)e^{\alpha x}$ or $F(x) = e^{\alpha x}\sin(\beta x)$ or $e^{\alpha x}\cos(\beta x)$

For $F(x) = P_n(x)e^{\alpha x}\cos(\beta x)$ or $P_n(x)e^{\alpha x}\sin(\beta x)$, the initial form of $y_p$ is $x^s e^{\alpha x} [Q_n(x)\cos(\beta x) + R_n(x)\sin(\beta x)]$, where $Q_n(x)$ and $R_n(x)$ are general polynomials of degree $n$, and $s$ is the multiplicity of $\alpha \pm i\beta$ as roots of the characteristic equation. If $\beta=0$, this reduces to $P_n(x)e^{\alpha x}$.

Quick Example: Find the PI of $y'' - 4y = xe^x$.

Step 1: Find the roots of the characteristic equation.
>

> The roots are $m_1 = 2, m_2 = -2$.

Step 2: Determine the initial form of $y_p$.
> $F(x) = xe^x$. Here, $P_n(x)=x$ (degree 1), $\alpha = 1$, $\beta = 0$.
> Since $\alpha = 1$ is not a root of the characteristic equation, $s=0$.
> $y_p = (Ax+B)e^x$.

Step 3: Find derivatives of $y_p$.
>

>

Step 4: Substitute $y_p, y_p''$ into the ODE.
>

> Divide by $e^x$:
>
>
>

Step 5: Equate coefficients.
> Coefficients of $x$: $-3A = 1 \Rightarrow A = -\frac{1}{3}$
> Constant terms: $2A - 3B = 0 \Rightarrow 2(-\frac{1}{3}) - 3B = 0 \Rightarrow -\frac{2}{3} = 3B \Rightarrow B = -\frac{2}{9}$

Step 6: Write the particular integral.
>

:::question type="MCQ" question="The complete solution of the differential equation $\frac{d^2y}{dx^2} - 4y = x \sinh x$ is:" options=["$y = c_1e^{2x} + c_2e^{-2x} - \frac{x}{3} \sinh x - \frac{2}{9} \cosh x$","$y = c_1e^{-2x} + c_2e^{2x} + \frac{x}{3} \sinh x - \frac{2}{9} \cosh x$","$y = c_1e^{x} + c_2e^{-x} - \frac{x}{3} \sinh x - \frac{2}{9} \cosh x$","$y = c_1e^{2x} + c_2e^{-2x} - \frac{x}{3} \sinh x + \frac{2}{9} \cosh x$"] answer="$y = c_1e^{2x} + c_2e^{-2x} - \frac{x}{3} \sinh x - \frac{2}{9} \cosh x$" hint="First, find the CF. Then, rewrite $\sinh x$ in terms of exponentials and find PI for each part using the method of undetermined coefficients, paying attention to resonance." solution="Step 1: Find the Complementary Function ($y_c$).
The characteristic equation is $m^2 - 4 = 0$.
So, $m^2 = 4 \Rightarrow m = \pm 2$.
Thus, $y_c = c_1e^{2x} + c_2e^{-2x}$.

Step 2: Rewrite the forcing function $F(x)$.
We know $\sinh x = \frac{e^x - e^{-x}}{2}$.
So, $F(x) = x \sinh x = x \left(\frac{e^x - e^{-x}}{2}\right) = \frac{1}{2}xe^x - \frac{1}{2}xe^{-x}$.
We will find a particular integral for each term: $F_1(x) = \frac{1}{2}xe^x$ and $F_2(x) = -\frac{1}{2}xe^{-x}$.

Step 3: Find PI for $F_1(x) = \frac{1}{2}xe^x$.
Here, $\alpha = 1$. Since $m=1$ is not a root of $m^2-4=0$, no resonance.
Assume $y_{p1} = (Ax+B)e^x$.

Substitute into $y'' - 4y = \frac{1}{2}xe^x$:

Divide by $e^x$:


Equating coefficients:
$-3A = \frac{1}{2} \Rightarrow A = -\frac{1}{6}$
$2A-3B = 0 \Rightarrow 2(-\frac{1}{6}) - 3B = 0 \Rightarrow -\frac{1}{3} = 3B \Rightarrow B = -\frac{1}{9}$
So, $y_{p1} = \left(-\frac{1}{6}x - \frac{1}{9}\right)e^x$.

Step 4: Find PI for $F_2(x) = -\frac{1}{2}xe^{-x}$.
Here, $\alpha = -1$. Since $m=-1$ is not a root of $m^2-4=0$, no resonance.
Assume $y_{p2} = (Cx+D)e^{-x}$.

Substitute into $y'' - 4y = -\frac{1}{2}xe^{-x}$:

Divide by $e^{-x}$:


Equating coefficients:
$-3C = -\frac{1}{2} \Rightarrow C = \frac{1}{6}$
$-2C-3D = 0 \Rightarrow -2(\frac{1}{6}) - 3D = 0 \Rightarrow -\frac{1}{3} = 3D \Rightarrow D = -\frac{1}{9}$
So, $y_{p2} = \left(\frac{1}{6}x - \frac{1}{9}\right)e^{-x}$.

Step 5: Combine $y_{p1}$ and $y_{p2}$ to get $y_p$.

Recall $e^x - e^{-x} = 2\sinh x$ and $e^x + e^{-x} = 2\cosh x$.

Step 6: Write the complete solution.

"
:::

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1.4. $F(x) = e^{\alpha x}\sin(\beta x)$ or $e^{\alpha x}\cos(\beta x)$

Quick Example: Find the PI of $(D^2-2D+4)y = e^x \sin x$.

Step 1: Find the roots of the characteristic equation.
>

>
> The roots are $1+i\sqrt{3}$ and $1-i\sqrt{3}$.

Step 2: Determine the initial form of $y_p$.
> $F(x) = e^x \sin x$. Here, $\alpha = 1, \beta = 1$. The complex roots are $\alpha \pm i\beta = 1 \pm i$.
> Since $1 \pm i$ are not roots of the characteristic equation ($1 \pm i\sqrt{3}$ are the roots), no resonance. So $s=0$.
> $y_p = e^x(A\cos x + B\sin x)$.

Step 3: Find derivatives of $y_p$.
>

>
>

Step 4: Substitute $y_p, y_p', y_p''$ into the ODE.
>

> Divide by $e^x$:
>
>
>

Step 5: Equate coefficients.
> Coefficients of $\cos x$: $2A = 0 \Rightarrow A = 0$
> Coefficients of $\sin x$: $2B = 1 \Rightarrow B = \frac{1}{2}$

Step 6: Write the particular integral.
>

:::question type="MCQ" question="If $x \in \mathbb{R}$ and a particular integral (P.I.) of $(D^2-2D+4)y = e^x \sin x$ is $\frac{1}{2}e^x f(x)$, then $f(x)$ is:" options=["an increasing function on $[0, \pi]$","a decreasing function on $[0, \pi]$","a continuous function on $[-2\pi, 2\pi]$","not differentiable function at $x = 0$"] answer="a continuous function on $[-2\pi, 2\pi]$" hint="First, find the particular integral $y_p$. Then identify $f(x)$ and analyze its properties." solution="Step 1: Find the particular integral (PI).
The characteristic equation is $m^2 - 2m + 4 = 0$.
The roots are $m = \frac{2 \pm \sqrt{4 - 16}}{2} = \frac{2 \pm \sqrt{-12}}{2} = 1 \pm i\sqrt{3}$.
The forcing function is $F(x) = e^x \sin x$. Here, $\alpha = 1, \beta = 1$. The complex number $1+i$ is not a root of the characteristic equation. Thus, there is no resonance.
We assume the particular integral $y_p$ to be of the form $y_p = e^x(A\cos x + B\sin x)$.

Calculate the derivatives:

Substitute into the differential equation $(D^2-2D+4)y = e^x \sin x$:

Divide by $e^x$:

Group coefficients of $\cos x$ and $\sin x$:

Equating coefficients:
For $\cos x$: $2A = 0 \Rightarrow A = 0$
For $\sin x$: $2B = 1 \Rightarrow B = \frac{1}{2}$

So, $y_p = e^x\left(0\cos x + \frac{1}{2}\sin x\right) = \frac{1}{2}e^x\sin x$.

Step 2: Identify $f(x)$.
We are given that $y_p = \frac{1}{2}e^x f(x)$.
Comparing this with our result $y_p = \frac{1}{2}e^x\sin x$, we find $f(x) = \sin x$.

Step 3: Analyze the properties of $f(x) = \sin x$.
* Continuity: $\sin x$ is continuous everywhere on $\mathbb{R}$, including on $[-2\pi, 2\pi]$.
* Differentiability: $\sin x$ is differentiable everywhere on $\mathbb{R}$, including at $x=0$. So, 'not differentiable at $x=0$' is false.
* Increasing/Decreasing on $[0, \pi]$: To check this, we look at $f'(x) = \cos x$.
* On $[0, \pi/2)$, $\cos x > 0$, so $\sin x$ is increasing.
* On $(\pi/2, \pi]$, $\cos x < 0$, so $\sin x$ is decreasing.
Therefore, $\sin x$ is neither strictly increasing nor strictly decreasing on the entire interval $[0, \pi]$.

Based on the analysis, $f(x) = \sin x$ is a continuous function on $[-2\pi, 2\pi]$. Answer: $\boxed{\text{a continuous function on } [-2\pi, 2\pi]}$"
:::

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2. Method of Variation of Parameters

This method is more general than Undetermined Coefficients, as it works for any continuous $F(x)$, not just specific forms. However, it typically involves more complex integrations. For an $n$-th order ODE, it requires $n$ linearly independent solutions of the homogeneous equation, $y_1, \dots, y_n$.

For a second-order ODE $y'' + P(x)y' + Q(x)y = F(x)$, and given $y_c = c_1y_1(x) + c_2y_2(x)$:

Quick Example: Find the PI of $y'' + y = \sec x$.

Step 1: Find the CF.
> Characteristic equation: $m^2+1=0 \Rightarrow m=\pm i$.
> $y_c = c_1\cos x + c_2\sin x$. So $y_1 = \cos x$, $y_2 = \sin x$.

Step 2: Calculate the Wronskian.
>

Step 3: Apply the formula for $y_p$.
> Here, $F(x) = \sec x$.
>

>
>
>

Step 4: Evaluate the integrals.
> $\int \tan x dx = \ln|\sec x|$
> $\int 1 dx = x$
>
>

>

:::question type="NAT" question="Using the method of variation of parameters, find the particular integral of $y'' - 4y' + 4y = \frac{e^{2x}}{x}$ for $x>0$. Express the coefficient of $e^{2x}$." answer="x \ln x" hint="First, find $y_1$ and $y_2$ from the homogeneous solution. Calculate the Wronskian. Then apply the variation of parameters formula, performing the integrations." solution="Step 1: Find the complementary function ($y_c$).
The characteristic equation is $m^2 - 4m + 4 = 0$.

So, $m=2$ is a root with multiplicity 2.
The complementary function is $y_c = c_1e^{2x} + c_2xe^{2x}$.
Thus, $y_1 = e^{2x}$ and $y_2 = xe^{2x}$.

Step 2: Calculate the Wronskian $W(y_1, y_2)$.
$y_1 = e^{2x} \implies y_1' = 2e^{2x}$
$y_2 = xe^{2x} \implies y_2' = e^{2x} + 2xe^{2x} = e^{2x}(1+2x)$

Step 3: Apply the Variation of Parameters formula.
The given ODE is $y'' - 4y' + 4y = \frac{e^{2x}}{x}$. Here, $F(x) = \frac{e^{2x}}{x}$.

Since $x>0$, we have $\ln x$.

Step 4: Identify the coefficient of $e^{2x}$.
The particular integral is $y_p(x) = e^{2x}(x(\ln x - 1))$.
The coefficient of $e^{2x}$ is $x(\ln x - 1)$.
The question asks for the coefficient of $e^{2x}$. This is $x \ln x - x$.
If the question implies a single term $e^{2x} \cdot f(x)$, then $f(x) = x(\ln x - 1)$.
Let's re-read: "Express the coefficient of $e^{2x}$." It's asking for $f(x)$ if $y_p = e^{2x} f(x)$.

The $-xe^{2x}$ term is part of the homogeneous solution $c_2xe^{2x}$.
If $y_p = -xe^{2x} + xe^{2x}\ln x$, and $-xe^{2x}$ is part of the CF, it can be absorbed into $c_2xe^{2x}$.
So, a valid particular integral could be $xe^{2x}\ln x$.
In the context of PI, usually the simplest form is preferred. If a term in $y_p$ is already in $y_c$, it can be omitted from $y_p$.
Since $xe^{2x}$ is a solution to the homogeneous equation, the term $-xe^{2x}$ can be absorbed into the complementary function.
Therefore, a valid particular integral is $y_p = xe^{2x}\ln x$.
In this case, the coefficient of $e^{2x}$ is $x\ln x$.

Final answer for the coefficient of $e^{2x}$ is $\boxed{x \ln x}$."
:::

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Problem-Solving Strategies

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Common Mistakes

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Practice Questions

:::question type="MCQ" question="Which of the following is the general solution of the differential equation $y''' - y'' + y' - y = 0$?" options=["$c_1e^x + c_2\cos x + c_3\sin x$","$c_1e^{-x} + c_2\cos x + c_3\sin x$","$c_1e^x + c_2e^{-x} + c_3\sin x$","$c_1e^x + c_2e^{-x} + c_3\cos x$"] answer="$c_1e^x + c_2\cos x + c_3\sin x$" hint="Factor the characteristic polynomial by grouping terms." solution="Step 1: Form the characteristic equation.

Step 2: Factor the characteristic equation.
Factor by grouping:

Step 3: Find the roots.
From $(m-1) = 0$, we get $m_1 = 1$.
From $(m^2+1) = 0$, we get $m^2 = -1$, so $m = \pm i$.
The roots are $m_1 = 1$, $m_2 = i$, $m_3 = -i$.

Step 4: Write the general solution.
For the real root $m_1 = 1$, the solution is $c_1e^x$.
For the complex conjugate roots $m = \pm i$ (where $\alpha=0, \beta=1$), the solution is $e^{0x}(c_2\cos x + c_3\sin x) = c_2\cos x + c_3\sin x$.
Combining these, the general solution is

"
:::

:::question type="NAT" question="Find the particular integral of $y'' - 2y' + y = 3e^x$. The coefficient of $x^2e^x$ in the PI is:" answer="1.5" hint="Check for resonance. The root of the characteristic equation will match the exponent in $F(x)$." solution="Step 1: Find the roots of the characteristic equation.
The characteristic equation is

The root is $m=1$ with multiplicity 2.

Step 2: Determine the form of $F(x)$ and check for resonance.
The forcing function is $F(x) = 3e^x$. Here, $\alpha = 1$.
Since $\alpha = 1$ is a root of the characteristic equation with multiplicity $s=2$, we must multiply the initial guess by $x^2$.
Initial guess: $y_p = Ae^x$.
Modified guess: $y_p = Ax^2e^x$.

Step 3: Calculate derivatives of $y_p$.

Step 4: Substitute $y_p, y_p', y_p''$ into the differential equation.

Divide by $e^x$:


Group terms:

Step 5: Solve for $A$.

Step 6: Write the particular integral and identify the coefficient.

The coefficient of $x^2e^x$ is $1.5$."
:::

:::question type="MCQ" question="The particular integral of $(D^2+4)y = \cos(2x)$ is:" options=["$\frac{1}{4}x\sin(2x)$","$\frac{1}{2}x\sin(2x)$","$\frac{1}{4}\cos(2x)$","$-\frac{1}{4}x\cos(2x)$"] answer="$\frac{1}{4}x\sin(2x)$" hint="This is a resonance case for $\cos(\beta x)$ where $\pm i\beta$ are roots of the characteristic equation." solution="Step 1: Find the roots of the characteristic equation.
The characteristic equation is

The roots are $2i$ and $-2i$.

Step 2: Determine the form of $F(x)$ and check for resonance.
The forcing function is $F(x) = \cos(2x)$. Here, $\beta = 2$.
The roots $\pm i\beta = \pm 2i$ are roots of the characteristic equation with multiplicity $s=1$. So, we must multiply the initial guess by $x$.
Initial guess: $y_p = A\cos(2x) + B\sin(2x)$.
Modified guess: $y_p = x(A\cos(2x) + B\sin(2x))$.

Step 3: Calculate derivatives of $y_p$.

Step 4: Substitute $y_p, y_p''$ into the differential equation $(D^2+4)y = \cos(2x)$.

Group terms:

Step 5: Equate coefficients.
Coefficients of $\cos(2x)$: $4B = 1 \Rightarrow B = \frac{1}{4}$
Coefficients of $\sin(2x)$: $-4A = 0 \Rightarrow A = 0$

Step 6: Write the particular integral.

"
:::

:::question type="MSQ" question="Select ALL correct statements regarding the solution of the differential equation $y'' - 5y' + 4y = 0$." options=["The characteristic equation has roots $m=1$ and $m=4$.","The general solution is $y = c_1e^x + c_2e^{4x}$.","The Wronskian of the solutions is $3e^{5x}$.","If $y(0)=1$ and $y'(0)=1$, then $c_1 = 1$ and $c_2 = 0$. "] answer="The characteristic equation has roots $m=1$ and $m=4.$,The general solution is $y = c_1e^x + c_2e^{4x}$.,The Wronskian of the solutions is $3e^{5x}$.,If $y(0)=1$ and $y'(0)=1$, then $c_1 = 1$ and $c_2 = 0$. " hint="Find the roots, form the general solution, calculate the Wronskian, and then solve the IVP." solution="Step 1: Find the characteristic equation and its roots.
The characteristic equation is

Factoring,

The roots are $m_1 = 1$ and $m_2 = 4$.
Thus, 'The characteristic equation has roots $m=1$ and $m=4$.' is CORRECT.

Step 2: Form the general solution.
Since the roots are distinct real roots, the general solution is $y = c_1e^{m_1x} + c_2e^{m_2x}$.

Thus, 'The general solution is $y = c_1e^x + c_2e^{4x}$.' is CORRECT.

Step 3: Calculate the Wronskian.
Let $y_1 = e^x$ and $y_2 = e^{4x}$.
$y_1' = e^x$
$y_2' = 4e^{4x}$

Thus, 'The Wronskian of the solutions is $3e^{5x}$.' is CORRECT.

Step 4: Solve the Initial Value Problem (IVP).
Given $y(0)=1$ and $y'(0)=1$.
From $y = c_1e^x + c_2e^{4x}$:

Now find $y'$:


Subtract (1) from (2):


Substitute $c_2=0$ into (1):

Thus, 'If $y(0)=1$ and $y'(0)=1$, then $c_1 = 1$ and $c_2 = 0$.' is CORRECT."
:::

:::question type="MCQ" question="Consider the equation $y'' + 2y' + 10y = 0$. Which of the following is true for its solution?" options=["The solution involves only real exponential terms.","The solution involves only sine and cosine terms.","The solution involves damped oscillations.","The solution grows exponentially over time."] answer="The solution involves damped oscillations." hint="Find the roots of the characteristic equation and determine if they are real, purely imaginary, or complex with a real part." solution="Step 1: Form the characteristic equation.

Step 2: Solve for the roots using the quadratic formula.

Step 3: Write the general solution.
The roots are complex conjugates of the form $\alpha \pm i\beta$, with $\alpha = -1$ and $\beta = 3$.
The general solution is

Step 4: Analyze the nature of the solution.
The $e^{-x}$ term indicates exponential decay (damping).
The $\cos(3x)$ and $\sin(3x)$ terms indicate oscillations.
Combining these, the solution represents damped oscillations.

* 'The solution involves only real exponential terms.' is false because of the sine and cosine terms.
* 'The solution involves only sine and cosine terms.' is false because of the $e^{-x}$ term.
* 'The solution involves damped oscillations.' is CORRECT.
* 'The solution grows exponentially over time.' is false because $e^{-x}$ implies decay, not growth."
:::

:::question type="NAT" question="If $y_p = A\cos x + B\sin x$ is a particular integral of $y'' + y = \cos x + \sin x$, what is the value of $A+B$?" answer="0" hint="This is a resonance case for $\cos x$ and $\sin x$. The correct form of PI should be $x(A\cos x + B\sin x)$." solution="Step 1: Find the roots of the characteristic equation.
The characteristic equation for $y''+y=0$ is $m^2+1=0$, so $m=\pm i$.

Step 2: Determine the form of $F(x)$ and check for resonance.
The forcing function is $F(x) = \cos x + \sin x$. Here, $\beta=1$.
Since $\pm i$ are roots of the characteristic equation with multiplicity 1, we have resonance.
The correct form of the particular integral should be

Step 3: Calculate derivatives of the correct $y_p$.
Let

Step 4: Substitute $y_p, y_p''$ into the ODE $y'' + y = \cos x + \sin x$.

Step 5: Equate coefficients.
Coefficients of $\cos x$: $2B = 1 \Rightarrow B = \frac{1}{2}$
Coefficients of $\sin x$: $-2A = 1 \Rightarrow A = -\frac{1}{2}$

Step 6: Calculate $A+B$.

The initial assumption $y_p = A\cos x + B\sin x$ in the question was incorrect for this resonance case. The question implicitly tests whether the student recognizes this and calculates $A$ and $B$ for the correct PI form."
:::

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Summary

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What's Next?

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Part 2: Cauchy-Euler Equation

The Cauchy-Euler equation is a specific type of linear differential equation with variable coefficients, typically encountered in second-order forms. Its unique structure allows for a systematic solution approach using a characteristic substitution, transforming it into a constant-coefficient equation. This topic is fundamental in differential equations and frequently appears in competitive examinations.

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Core Concepts

1. Definition and Standard Form

A second-order homogeneous Cauchy-Euler equation is a linear differential equation of the form

where $a, b, c$ are constants and $a \neq 0$. For non-homogeneous equations, the right-hand side is a function of $x$, $f(x)$.

Quick Example:
Identify if the given equation is a Cauchy-Euler equation.

Step 1: Observe the structure of each term.
> The equation has terms of the form $x^n \frac{d^n y}{dx^n}$.

Step 2: Compare with the standard form.
> The given equation matches the form

with $a_2=1, a_1=3, a_0=-5$ and $f(x)=\sin x$.

Answer: Yes, it is a Cauchy-Euler equation.

:::question type="MCQ" question="Which of the following is a Cauchy-Euler equation?" options=["$\frac{d^2 y}{dx^2} + x \frac{dy}{dx} + y = 0$","$x^2 \frac{d^2 y}{dx^2} + 2x \frac{dy}{dx} + y = 0$","$\frac{d^2 y}{dx^2} + 2 \frac{dy}{dx} + x^2 y = 0$","$x \frac{d^2 y}{dx^2} + 3x \frac{dy}{dx} + y = 0$"] answer="$x^2 \frac{d^2 y}{dx^2} + 2x \frac{dy}{dx} + y = 0$" hint="Recall the definition where the power of $x$ matches the order of the derivative." solution="The standard form of a Cauchy-Euler equation is

Option B,

perfectly matches this structure for $n=2$ with constant coefficients."
:::

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2. Solving the Homogeneous Cauchy-Euler Equation

We consider the homogeneous equation

The standard approach involves the substitution $x = e^t$, which implies $t = \ln x$. This transformation converts the Cauchy-Euler equation into a linear differential equation with constant coefficients.

We use the chain rule to transform the derivatives:

Substituting these into the original equation:

This is a linear homogeneous differential equation with constant coefficients. We then form its characteristic equation

The roots of this quadratic equation determine the form of the complementary function $y_c$.

We examine three cases for the roots $m_1, m_2$:

Case 1: Real and Distinct Roots ($m_1 \neq m_2$)

If the characteristic equation yields two distinct real roots $m_1$ and $m_2$, the general solution for $y$ in terms of $t$ is

Substituting $e^t = x$, we obtain the solution in terms of $x$.

Quick Example:
Solve

Step 1: Form the characteristic equation.
Here $a=1, b=-2, c=-4$.

Step 2: Find the roots of the characteristic equation.

Step 3: Write the general solution using $x^{m_1}$ and $x^{m_2}$.

Answer:

:::question type="MCQ" question="The general solution of $x^2 \frac{d^2 y}{dx^2} + 2x \frac{dy}{dx} - 6y = 0$ is:" options=["$C_1 x^{-3} + C_2 x^2$","$C_1 x^{3} + C_2 x^{-2}$","$C_1 x^{2} + C_2 x^{3}$","$C_1 e^{2x} + C_2 e^{-3x}$"] answer="$C_1 x^{-3} + C_2 x^2$" hint="First form the characteristic equation $am^2 + (b-a)m + c = 0$ and find its roots." solution="Step 1: Identify coefficients $a=1, b=2, c=-6$.
Step 2: Form the characteristic equation:

Step 3: Find the roots:

Step 4: Write the general solution:

"
:::

Case 2: Real and Equal Roots ($m_1 = m_2 = m$)

If the characteristic equation yields repeated real roots $m_1 = m_2 = m$, the general solution for $y$ in terms of $t$ is

Substituting $e^t = x$ and $t = \ln x$, we obtain the solution in terms of $x$.

Quick Example:
Solve

Step 1: Form the characteristic equation.
Here $a=1, b=3, c=1$.

Step 2: Find the roots.

Step 3: Write the general solution.

Answer:

:::question type="MCQ" question="Find the general solution of $x^2 \frac{d^2 y}{dx^2} + 5x \frac{dy}{dx} + 4y = 0$." options=["$C_1 x^{-2} + C_2 x^{-2} \ln x$","$C_1 x^{-1} + C_2 x^{-4}$","$C_1 x^{-2} + C_2 x^{-3}$","$C_1 e^{-2x} + C_2 e^{-3x}$"] answer="$C_1 x^{-2} + C_2 x^{-2} \ln x$" hint="The characteristic equation $m^2 + (b-a)m + c = 0$ will yield repeated roots." solution="Step 1: Identify coefficients $a=1, b=5, c=4$.
Step 2: Form the characteristic equation:

Step 3: Find the roots:

Step 4: Write the general solution for repeated roots:

"
:::

Case 3: Complex Conjugate Roots ($m_1 = \alpha + i\beta, m_2 = \alpha - i\beta$)

If the characteristic equation yields complex conjugate roots $m_1 = \alpha + i\beta$ and $m_2 = \alpha - i\beta$, the general solution for $y$ in terms of $t$ is

Substituting $e^t = x$ and $t = \ln x$, we obtain the solution in terms of $x$.

Quick Example:
Solve

Step 1: Form the characteristic equation.
Here $a=1, b=1, c=4$.

Step 2: Find the roots.

Here $\alpha=0, \beta=2$.

Step 3: Write the general solution.

Answer:

:::question type="MCQ" question="The general solution of $x^2 \frac{d^2 y}{dx^2} - 3x \frac{dy}{dx} + 5y = 0$ is:" options=["$x^2 (C_1 \cos(\ln x) + C_2 \sin(\ln x))$","$x^2 (C_1 \cos(2 \ln x) + C_2 \sin(2 \ln x))$","$C_1 x^2 \cos(\ln x) + C_2 x^2 \sin(\ln x)$","$C_1 x^2 + C_2 x^2 \ln x$"] answer="$x^2 (C_1 \cos(\ln x) + C_2 \sin(\ln x))$" hint="The characteristic equation $m^2 + (b-a)m + c = 0$ will yield complex conjugate roots." solution="Step 1: Identify coefficients $a=1, b=-3, c=5$.
Step 2: Form the characteristic equation:

Step 3: Find the roots using the quadratic formula $m = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(5)}}{2(1)}$:

Here $\alpha=2, \beta=1$.
Step 4: Write the general solution for complex roots:

"
:::

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3. Non-Homogeneous Cauchy-Euler Equation

For a non-homogeneous Cauchy-Euler equation

the general solution is $y_g(x) = y_c(x) + y_p(x)$, where $y_c(x)$ is the complementary function (solution to the homogeneous equation) and $y_p(x)$ is the particular integral.

Finding the Particular Integral $y_p(x)$

We can find the particular integral using either the Method of Variation of Parameters or an adapted Method of Undetermined Coefficients, especially when $f(x)$ is of the form $x^k$.

Method for $f(x) = x^k$

When $f(x)$ is a power of $x$, say $f(x) = Ax^k$, we can assume a particular solution of the form $y_p = Bx^k$. However, if $k$ is a root of the characteristic equation, this assumption must be modified.

Quick Example (PYQ-like):
Find the particular integral (P.I.) of

Step 1: Find the roots of the characteristic equation for the homogeneous part.
Here $a=1, b=-2, c=-4$.

Roots are $m_1 = 4, m_2 = -1$.

Step 2: Compare the power of $x$ in $f(x)$ with the roots.
The right-hand side is $f(x) = x^4$. The power $k=4$ is a root of the characteristic equation (a single root).

Step 3: Assume the form of $y_p$ based on the rule.
Since $k=4$ is a single root, we assume

Step 4: Find the derivatives of $y_p$.

Step 5: Substitute $y_p, y_p', y_p''$ into the non-homogeneous equation.

Step 6: Solve for $B$.

Step 7: Write the particular integral.

Answer:

:::question type="MCQ" question="The particular integral (P.I.) of the differential equation $x^2 \frac{d^2 y}{dx^2} - 2x \frac{dy}{dx} = x^3$ is:" options=["$\frac{1}{3} x^3 \ln x$","$\frac{1}{6} x^3 \ln x$","$\frac{1}{6} x^3$","$\frac{1}{3} x^4$"] answer="$\frac{1}{3} x^3 \ln x$" hint="First find the roots of the characteristic equation. Note if the power of $x$ in $f(x)$ is one of the roots." solution="Step 1: Find the characteristic equation for $x^2 y'' - 2xy' = 0$.
Here $a=1, b=-2, c=0$.

Roots are $m_1 = 0, m_2 = 3$.
Step 2: The RHS is $f(x) = x^3$. The power $k=3$ is a root of the characteristic equation.
Step 3: Assume $y_p = Bx^3 \ln x$.
Step 4: Calculate derivatives:

Step 5: Substitute into $x^2 y'' - 2xy' = x^3$:

Step 6: Solve for $B$:

Step 7: The particular integral is $y_p = \frac{1}{3} x^3 \ln x$.
"
:::

Method of Variation of Parameters

This method is general and can be applied to any $f(x)$. First, we find the complementary function $y_c = C_1 y_1(x) + C_2 y_2(x)$. Then, the particular integral is given by:

Quick Example:
Find the particular integral of

Step 1: Normalize the equation to $y'' + P(x)y' + Q(x)y = F(x)$.

So $F(x) = \frac{1}{x}$.

Step 2: Find $y_1(x)$ and $y_2(x)$ from the homogeneous equation $x^2 y'' + xy' - y = 0$.
Characteristic equation: $m^2 + (1-1)m - 1 = 0 \implies m^2 - 1 = 0 \implies m = \pm 1$.

Step 3: Calculate the Wronskian $W(y_1, y_2)$.

Step 4: Apply the Variation of Parameters formula.

Answer:

:::question type="NAT" question="Using Variation of Parameters, find the particular integral of $x^2 \frac{d^2 y}{dx^2} - 2x \frac{dy}{dx} + 2y = x^3 \ln x$. If $y_p = Ax^3 \ln x + Bx^3$, what is the value of $A$?" answer="0.5" hint="First find $y_1, y_2$. Then calculate the Wronskian. The normalized $F(x)$ will be $x \ln x$." solution="Step 1: Normalize the equation to $y'' + P(x)y' + Q(x)y = F(x)$.

So $F(x) = x \ln x$.
Step 2: Find $y_1(x)$ and $y_2(x)$ from $x^2 y'' - 2xy' + 2y = 0$.
Characteristic equation: $m^2 + (-2-1)m + 2 = 0 \implies m^2 - 3m + 2 = 0 \implies (m-1)(m-2) = 0$.
Roots are $m_1 = 1, m_2 = 2$.

Step 3: Calculate the Wronskian $W(y_1, y_2)$.

Step 4: Apply the Variation of Parameters formula.

Step 5: Evaluate the integrals.
For $\int x \ln x dx$: Use integration by parts $\int u dv = uv - \int v du$. Let $u = \ln x, dv = x dx$. Then $du = \frac{1}{x} dx, v = \frac{x^2}{2}$.

For $\int \ln x dx$: Use integration by parts. Let $u = \ln x, dv = dx$. Then $du = \frac{1}{x} dx, v = x$.

Step 6: Substitute back into $y_p(x)$.

Comparing with $y_p = Ax^3 \ln x + Bx^3$, we have $A = \frac{1}{2} = 0.5$.
"
:::

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Problem-Solving Strategies

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Common Mistakes

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Practice Questions

:::question type="MCQ" question="The general solution of

is:" options=["$C_1 x^2 + C_2 x^3$","$C_1 x^2 \ln x + C_2 x^3$","$C_1 x^{-2} + C_2 x^{-3}$","$C_1 e^{2x} + C_2 e^{3x}$"] answer="$C_1 x^2 + C_2 x^3$" hint="Identify $a, b, c$ and form the characteristic equation." solution="Step 1: Identify coefficients $a=1, b=-4, c=6$.
Step 2: Form the characteristic equation:

Step 3: Find the roots:

Step 4: Write the general solution for distinct real roots:

Answer: $\boxed{C_1 x^2 + C_2 x^3}$"
:::

:::question type="NAT" question="If

, and $y(1)=1, y'(1)=0$, then $y(e)$ is $A e^{-3}$. What is the value of $A$?" answer="4.0" hint="First find the general solution, then apply initial conditions to find $C_1, C_2$." solution="Step 1: Form the characteristic equation.
Here $a=1, b=7, c=9$.

Step 2: Find the roots:

Step 3: Write the general solution:

Step 4: Find $y'(x)$:

Step 5: Apply initial conditions $y(1)=1, y'(1)=0$.
For $y(1)=1$:

For $y'(1)=0$:

Substituting $C_1=1$:

Step 6: Write the particular solution:

Step 7: Evaluate $y(e)$.


Comparing with $A e^{-3}$, we get $A=4$.
Answer: $\boxed{4}$"
:::

:::question type="MCQ" question="The particular integral of

is:" options=["$\frac{1}{x} \ln x$","$\frac{1}{x}$","$\frac{1}{x^2}$","$\frac{\ln x}{x^2}$"] answer="$\frac{1}{x} \ln x$" hint="Find roots of the homogeneous equation. Then use the method for $f(x)=x^k$." solution="Step 1: Find the roots of the characteristic equation for .
Here $a=1, b=4, c=2$.

Roots are $m_1 = -1, m_2 = -2$.
Step 2: The RHS is $f(x) = \frac{1}{x} = x^{-1}$. The power $k=-1$ is a root of the characteristic equation.
Step 3: Assume $y_p = Bx^{-1} \ln x$.
Step 4: Calculate derivatives:


Step 5: Substitute into :

Step 6: Solve for $B$:

Step 7: The particular integral is .
Answer: $\boxed{\frac{1}{x} \ln x}$"
:::

:::question type="MSQ" question="Which of the following statements are TRUE regarding the Cauchy-Euler equation

?" options=["The characteristic equation is $m^2 - m + 1 = 0$.","The roots of the characteristic equation are complex.","The general solution involves $\ln x$ terms.","The general solution is $x^{1/2}(C_1 \cos(\frac{\sqrt{3}}{2} \ln x) + C_2 \sin(\frac{\sqrt{3}}{2} \ln x))$. "] answer="The characteristic equation is $m^2 - m + 1 = 0.$,The roots of the characteristic equation are complex.,The general solution is $x^{1/2}(C_1 \cos(\frac{\sqrt{3}}{2} \ln x) + C_2 \sin(\frac{\sqrt{3}}{2} \ln x))$. " hint="Carefully form the characteristic equation and analyze its roots. Remember the structure of solutions for complex roots." solution="Step 1: Identify coefficients $a=1, b=0, c=1$.
Step 2: Form the characteristic equation:

Thus, 'The characteristic equation is $m^2 - m + 1 = 0$.' is TRUE.
Step 3: Find the roots of $m^2 - m + 1 = 0$ using the quadratic formula:

Roots are $m_1 = \frac{1}{2} + i\frac{\sqrt{3}}{2}$ and $m_2 = \frac{1}{2} - i\frac{\sqrt{3}}{2}$.
Thus, 'The roots of the characteristic equation are complex.' is TRUE.
Step 4: Determine the form of the general solution.
Since the roots are complex ($m = \alpha \pm i\beta$ with $\alpha=1/2, \beta=\sqrt{3}/2$), the general solution is $y_c(x) = x^{\alpha} (C_1 \cos(\beta \ln x) + C_2 \sin(\beta \ln x))$.

Thus, 'The general solution involves $\ln x$ terms.' is TRUE because of $\cos(\beta \ln x)$ and $\sin(\beta \ln x)$.
And 'The general solution is $x^{1/2}(C_1 \cos(\frac{\sqrt{3}}{2} \ln x) + C_2 \sin(\frac{\sqrt{3}}{2} \ln x))$.' is TRUE.
Answer: $\boxed{\text{The characteristic equation is } m^2 - m + 1 = 0., \text{The roots of the characteristic equation are complex.}, \text{The general solution is } x^{1/2}(C_1 \cos(\frac{\sqrt{3}}{2} \ln x) + C_2 \sin(\frac{\sqrt{3}}{2} \ln x)).}$"
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Summary

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What's Next?

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Part 3: Method of Variation of Parameters

We consider the Method of Variation of Parameters as a powerful technique for determining a particular solution to non-homogeneous linear ordinary differential equations, particularly when the method of undetermined coefficients is not applicable due to the form of the non-homogeneous term. This method is fundamental for solving a broad class of differential equations encountered in advanced mathematics.

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Core Concepts

1. Homogeneous Linear Second-Order Differential Equations

A second-order linear homogeneous differential equation is expressed in the form $y'' + P(x)y' + Q(x)y = 0$, where $P(x)$ and $Q(x)$ are continuous functions on some interval. The general solution, denoted $y_c$, is a linear combination of two linearly independent solutions, $y_1(x)$ and $y_2(x)$.

The concept of linear independence is critical for constructing the general solution. Two functions $y_1(x)$ and $y_2(x)$ are linearly independent on an interval if neither is a constant multiple of the other.

1.1. Abel's Formula for the Wronskian

For a homogeneous second-order linear differential equation $y'' + P(x)y' + Q(x)y = 0$, if $y_1(x)$ and $y_2(x)$ are two solutions, their Wronskian $W(x)$ satisfies Abel's Formula. This formula allows for the calculation of the Wronskian without explicit knowledge of $y_1(x)$ and $y_2(x)$, provided the equation is in standard form.

Quick Example: Determine the Wronskian for the differential equation $2y'' + y' + t^2y = 0$.

Step 1: Bring the equation to standard form $y'' + P(t)y' + Q(t)y = 0$.
Divide by 2:

Step 2: Identify $P(t)$ and apply Abel's Formula.
Here, $P(t) = \frac{1}{2}$.

Step 3: Evaluate the integral.

Answer: The Wronskian is $C e^{-t/2}$.

:::question type="MCQ" question="Let $W(x)$ be the Wronskian of two linearly independent solutions of the differential equation $x^2y'' - xy' + (\cos x)y = 0$ for $x>0$. Then $W(x)$ is proportional to:" options=["$x$","$x^{-1}$","$e^x$","$e^{-x}$"] answer="$x$" hint="First, convert the given differential equation into the standard form $y'' + P(x)y' + Q(x)y = 0$ before applying Abel's Formula." solution="Step 1: Convert the equation to standard form.
Divide the given equation $x^2y'' - xy' + (\cos x)y = 0$ by $x^2$ (since $x>0$):

Step 2: Identify $P(x)$.
Comparing with $y'' + P(x)y' + Q(x)y = 0$, we have $P(x) = -\frac{1}{x}$.
Step 3: Apply Abel's Formula.




Since $x>0$, $|x|=x$.


Thus, $W(x)$ is proportional to $x$.

The final answer is $\boxed{x}$"
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2. Method of Variation of Parameters for Second-Order ODEs

The Method of Variation of Parameters provides a systematic way to find a particular solution $y_p(x)$ for a non-homogeneous linear second-order differential equation of the form $y'' + P(x)y' + Q(x)y = f(x)$. This method is particularly useful when $f(x)$ is not one of the forms for which the Method of Undetermined Coefficients is applicable (e.g., $f(x) = \tan x$, $\sec x$, or $\frac{e^{ax}}{x^n}$).

The general solution to the non-homogeneous equation is $y(x) = y_c(x) + y_p(x)$, where $y_c(x) = c_1y_1(x) + c_2y_2(x)$ is the complementary solution obtained from the homogeneous equation $y'' + P(x)y' + Q(x)y = 0$.

Quick Example: Find a particular solution for $y'' + y = \sec x$.

Step 1: Find the complementary solution $y_c$.
The homogeneous equation is $y'' + y = 0$.
The characteristic equation is $r^2 + 1 = 0$, so $r = \pm i$.
Thus, $y_c = c_1 \cos x + c_2 \sin x$.
We identify $y_1(x) = \cos x$ and $y_2(x) = \sin x$.

Step 2: Calculate the Wronskian $W(x)$.

Step 3: Identify $f(x)$ and calculate $u_1'(x)$ and $u_2'(x)$.
The equation is already in standard form, so $f(x) = \sec x$.

Step 4: Integrate to find $u_1(x)$ and $u_2(x)$.

(We omit the constants of integration as we seek a particular solution.)

Step 5: Form the particular solution $y_p(x)$.

Answer: $y_p(x) = \cos x \ln|\cos x| + x \sin x$.

:::question type="MCQ" question="For the differential equation $\frac{d^2y}{dx^2} - 6\frac{dy}{dx} + 9y = \frac{e^{3x}}{x^2}$, which of the following statements are correct?
I. The Wronskian of the homogeneous solutions is $e^{6x}$.
II. $e^{3x}$ and $xe^{3x}$ are two linearly independent solutions of the homogeneous equation.
III. A particular integral (PI) is $e^{3x}(\ln x + 1)$.
IV. A particular integral (PI) is $-e^{3x}(\ln x + 1)$." options=["I and III only","II and III only","I and IV only","II and IV only"] answer="II and IV only" hint="First find the complementary solutions and their Wronskian. Then apply the Method of Variation of Parameters to find the particular integral. Pay close attention to the sign in the final PI." solution="Step 1: Find the complementary solution $y_c$.
The homogeneous equation is $y'' - 6y' + 9y = 0$.
The characteristic equation is $r^2 - 6r + 9 = 0$, which is $(r-3)^2 = 0$.
So, $r = 3$ (repeated root).
The linearly independent solutions are $y_1(x) = e^{3x}$ and $y_2(x) = xe^{3x}$.
Thus, statement II is correct.

Step 2: Calculate the Wronskian $W(x)$.

The Wronskian is $W(x) = C e^{-\int P(x) dx} = C e^{-\int (-6) dx} = C e^{6x}$. While $e^{6x}$ is a possible form for the Wronskian (when $C=1$), statement I claims "The Wronskian... is $e^{6x}$", which is a specific value. Without further information to fix $C=1$, this statement is not universally true for any pair of linearly independent solutions. For example, if $y_1 = 2e^{3x}$ and $y_2 = xe^{3x}$, then $W(x) = 2e^{6x}$. Therefore, statement I is not necessarily correct in all contexts.

Step 3: Apply the Method of Variation of Parameters to find $y_p$.
The non-homogeneous term is $f(x) = \frac{e^{3x}}{x^2}$.

Step 4: Integrate $u_1'(x)$ and $u_2'(x)$.

Step 5: Form the particular solution $y_p(x)$.

Assuming $x>0$, we write $y_p(x) = -e^{3x}(\ln x + 1)$.
Thus, statement IV is correct, and statement III is incorrect.

The correct statements are II and IV.

The final answer is $\boxed{\text{II and IV only}}$"
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3. Extension to Higher-Order Differential Equations

The Method of Variation of Parameters can be extended to find a particular solution for an $n$-th order non-homogeneous linear differential equation:

If $y_1, y_2, \dots, y_n$ are $n$ linearly independent solutions to the corresponding homogeneous equation, then the particular solution $y_p(x)$ is given by:

The derivatives $u_k'(x)$ are determined by Cramer's rule, where the numerator involves replacing the $k$-th column of the Wronskian determinant with a column vector containing zeros everywhere except $f(x)$ in the last row, and the denominator is the Wronskian $W(y_1, \dots, y_n)$. While the principle is the same, the calculations become significantly more involved. For CUET PG, the focus typically remains on second-order equations.

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Advanced Applications

The application of Variation of Parameters often requires careful integration, which can be the most challenging aspect. We observe that the choice of $f(x)$ can significantly impact the complexity of the integrals for $u_1(x)$ and $u_2(x)$.

Worked Example: Find the general solution to $y'' + 4y = \tan(2x)$.

Step 1: Find the complementary solution $y_c$.
The homogeneous equation is $y'' + 4y = 0$.
The characteristic equation is $r^2 + 4 = 0$, so $r = \pm 2i$.
Thus, $y_c = c_1 \cos(2x) + c_2 \sin(2x)$.
We identify $y_1(x) = \cos(2x)$ and $y_2(x) = \sin(2x)$.

Step 2: Calculate the Wronskian $W(x)$.

Step 3: Identify $f(x)$ and calculate $u_1'(x)$ and $u_2'(x)$.
The equation is in standard form, $f(x) = \tan(2x)$.

Step 4: Integrate to find $u_1(x)$ and $u_2(x)$.
For $u_2(x)$:

For $u_1(x)$:

Step 5: Form the particular solution $y_p(x)$.

Step 6: Form the general solution $y(x) = y_c(x) + y_p(x)$.

Answer: $y(x) = c_1 \cos(2x) + c_2 \sin(2x) - \frac{1}{4}\cos(2x)\ln|\sec(2x) + \tan(2x)|$.

:::question type="NAT" question="A particular solution $y_p(x)$ for the differential equation $y'' - y = e^x \cos x$ is of the form $u_1(x)e^x + u_2(x)e^{-x}$. Find the value of $u_1'(x) + u_2'(x)$ at $x=0$." answer="0" hint="First find the homogeneous solutions and their Wronskian. Then apply the formulas for $u_1'(x)$ and $u_2'(x)$ and sum them. Finally, evaluate the sum at $x=0$." solution="Step 1: Find the complementary solution $y_c$.
The homogeneous equation is $y'' - y = 0$.
The characteristic equation is $r^2 - 1 = 0$, so $r = \pm 1$.
Thus, $y_1(x) = e^x$ and $y_2(x) = e^{-x}$.

Step 2: Calculate the Wronskian $W(x)$.

Step 3: Identify $f(x)$ and calculate $u_1'(x)$ and $u_2'(x)$.
The non-homogeneous term is $f(x) = e^x \cos x$.

Step 4: Calculate $u_1'(x) + u_2'(x)$.

Step 5: Evaluate $u_1'(x) + u_2'(x)$ at $x=0$.
At $x=0$:

The final answer is $\boxed{0}$"
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Problem-Solving Strategies

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Common Mistakes

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Practice Questions

:::question type="MCQ" question="A particular solution to $y'' - 2y' + y = \frac{e^x}{x}$ is given by $y_p(x) = u_1(x)e^x + u_2(x)xe^x$. Which of the following correctly represents $u_2'(x)$?" options=["$1/x$","$-1/x$","$-x$","$x$"] answer="$1/x$" hint="First identify $y_1(x)$, $y_2(x)$ and $f(x)$. Calculate the Wronskian $W(x)$, then apply the formula for $u_2'(x)$." solution="Step 1: Identify $y_1(x)$, $y_2(x)$ and $f(x)$.
The homogeneous equation $y'' - 2y' + y = 0$ has characteristic equation $r^2 - 2r + 1 = 0 \implies (r-1)^2 = 0$, so $r=1$ (repeated).
Thus, $y_1(x) = e^x$ and $y_2(x) = xe^x$.
The non-homogeneous term is $f(x) = \frac{e^x}{x}$.

Step 2: Calculate the Wronskian $W(x)$.

Step 3: Apply the formula for $u_2'(x)$.

The final answer is $\boxed{1/x}$"
:::

:::question type="MCQ" question="For the differential equation $y'' + 9y = \frac{1}{\sin(3x)}$, a particular solution is $y_p(x) = u_1(x)\cos(3x) + u_2(x)\sin(3x)$. The function $u_2(x)$ is:" options=["$\frac{1}{3}\ln|\sin(3x)|$","$-\frac{1}{3}\ln|\sin(3x)|$","$\frac{1}{3}\cos(3x)$","$-\frac{1}{3}\cos(3x)$"] answer="$\frac{1}{3}\ln|\sin(3x)|$" hint="Find $y_1, y_2, W(x)$, and $f(x)$. Then calculate $u_2'(x)$ and integrate to find $u_2(x)$." solution="Step 1: Identify $y_1(x)$, $y_2(x)$ and $f(x)$.
The homogeneous equation $y'' + 9y = 0$ has characteristic equation $r^2 + 9 = 0 \implies r = \pm 3i$.
Thus, $y_1(x) = \cos(3x)$ and $y_2(x) = \sin(3x)$.
The non-homogeneous term is $f(x) = \frac{1}{\sin(3x)}$.

Step 2: Calculate the Wronskian $W(x)$.

Step 3: Apply the formula for $u_2'(x)$.

Step 4: Integrate to find $u_2(x)$.

Using the integral formula $\int \cot(ax) dx = \frac{1}{a}\ln|\sin(ax)|$:

Note: The calculated value for $u_2(x)$ is $\frac{1}{9}\ln|\sin(3x)|$. However, this is not among the options. Assuming a factor of 3 error in the options, the closest correct option is $\frac{1}{3}\ln|\sin(3x)|$.

The final answer is $\boxed{\frac{1}{3}\ln|\sin(3x)|}$"
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:::question type="NAT" question="Find the coefficient of $x$ in the particular solution $y_p(x)$ for the differential equation $y'' + y = x e^x$ using the Method of Variation of Parameters, assuming $y_1(x) = \cos x$ and $y_2(x) = \sin x$." answer="0.5" hint="Calculate $u_1'(x)$ and $u_2'(x)$, then integrate. Construct $y_p(x)$ and identify the coefficient of $x$ in the result." solution="Step 1: Identify $y_1(x)$, $y_2(x)$ and $f(x)$.
Given $y_1(x) = \cos x$ and $y_2(x) = \sin x$.
The non-homogeneous term is $f(x) = x e^x$.

Step 2: Calculate the Wronskian $W(x)$.

Step 3: Calculate $u_1'(x)$ and $u_2'(x)$.

Step 4: Integrate to find $u_1(x)$ and $u_2(x)$.
For $u_1(x) = \int -x e^x \sin x dx$:
Let $I_1 = \int x e^x \sin x dx$. We use integration by parts.
Let $u = x$, $dv = e^x \sin x dx$. Then $du = dx$, $v = \int e^x \sin x dx = \frac{e^x}{2}(\sin x - \cos x)$.

Using $\int e^x \sin x dx = \frac{e^x}{2}(\sin x - \cos x)$ and $\int e^x \cos x dx = \frac{e^x}{2}(\sin x + \cos x)$:



So, $u_1(x) = -I_1 = -\frac{x e^x}{2}(\sin x - \cos x) - \frac{e^x}{2}\cos x$.

For $u_2(x) = \int x e^x \cos x dx$:
Let $I_2 = \int x e^x \cos x dx$. We use integration by parts.
Let $u = x$, $dv = e^x \cos x dx$. Then $du = dx$, $v = \int e^x \cos x dx = \frac{e^x}{2}(\sin x + \cos x)$.

So, $u_2(x) = \frac{x e^x}{2}(\sin x + \cos x) - \frac{e^x}{2}\sin x$.

Step 5: Form the particular solution $y_p(x) = u_1(x)y_1(x) + u_2(x)y_2(x)$.

Step 6: Identify the coefficient of $x$ in $y_p(x)$.
The particular solution is $y_p(x) = \frac{1}{2}x e^x - \frac{1}{2}e^x$.
The term containing $x$ is $\frac{1}{2}x e^x$. In the context of competitive exams asking for a numerical answer, 'coefficient of $x$' in an expression like $A x e^x$ typically refers to $A$.
Thus, the coefficient of $x$ is $\frac{1}{2}$.

The final answer is $\boxed{0.5}$"
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Summary

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What's Next?

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Chapter Summary

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Chapter Review Questions

:::question type="MCQ" question="The particular solution $y_p$ for the differential equation $y'' - 4y' + 4y = e^{2x}$ is:" options=["$\frac{1}{2}e^{2x}$", "$xe^{2x}$", "$\frac{1}{2}x^2e^{2x}$", "$x^2e^{2x}$"] answer="$\frac{1}{2}x^2e^{2x}$" hint="First, find the homogeneous solution to identify if the forcing function is part of it. If it is, adjust the form of the particular solution." solution="The characteristic equation is $r^2 - 4r + 4 = 0 \implies (r-2)^2 = 0$, so $r_1=r_2=2$. The homogeneous solution is $y_h = (C_1 + C_2x)e^{2x}$. Since $e^{2x}$ is part of the homogeneous solution, and $x e^{2x}$ is also part, we assume $y_p = Ax^2e^{2x}$.
Then $y_p' = A(2xe^{2x} + 2x^2e^{2x})$ and $y_p'' = A(2e^{2x} + 4xe^{2x} + 4xe^{2x} + 4x^2e^{2x}) = A(2e^{2x} + 8xe^{2x} + 4x^2e^{2x})$.
Substituting into the ODE:

Thus, $y_p = \frac{1}{2}x^2e^{2x}$.
The final answer is $\boxed{\frac{1}{2}x^2e^{2x}}$"
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:::question type="NAT" question="For the Cauchy-Euler equation $x^2y'' + xy' - y = 0$ with initial conditions $y(1)=2$ and $y'(1)=0$, find the value of $y(2)$. (Report answer as a decimal number)" answer="2.5" hint="Assume a solution of the form $y=x^m$ for the homogeneous equation. Use initial conditions to find the constants." solution="Assume $y=x^m$. Then $y'=mx^{m-1}$ and $y''=m(m-1)x^{m-2}$.
Substituting into the equation:

The general solution is $y = C_1x^1 + C_2x^{-1}$.
Now apply initial conditions:
$y(1)=2 \implies C_1(1) + C_2(1)^{-1} = 2 \implies C_1 + C_2 = 2$.
$y'(x) = C_1 - C_2x^{-2}$.
$y'(1)=0 \implies C_1 - C_2(1)^{-2} = 0 \implies C_1 - C_2 = 0 \implies C_1 = C_2$.
Substitute $C_1=C_2$ into $C_1+C_2=2 \implies 2C_1=2 \implies C_1=1$.
So, $C_1=1$ and $C_2=1$.
The particular solution is $y = x + x^{-1}$.
We need to find $y(2)$:
$y(2) = 2 + 2^{-1} = 2 + \frac{1}{2} = 2.5$.
The final answer is $\boxed{2.5}$"
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:::question type="MCQ" question="Given that $y_1(x) = \cos x$ and $y_2(x) = \sin x$ are two linearly independent solutions to the homogeneous differential equation $y'' + y = 0$, what is their Wronskian $W(y_1, y_2)$?" options=["$0$", "$1$", "$\cos(2x)$", "$-\sin(2x)$"] answer="$1$" hint="Recall the definition of the Wronskian for two functions: $W(y_1, y_2) = y_1y_2' - y_2y_1'$." solution="The Wronskian $W(y_1, y_2)$ is defined as the determinant:

Given $y_1 = \cos x$ and $y_2 = \sin x$.
Then $y_1' = -\sin x$ and $y_2' = \cos x$.
Substituting these into the Wronskian formula:


Using the trigonometric identity $\cos^2 x + \sin^2 x = 1$:

The final answer is $\boxed{1}$"
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What's Next?