Solve the differential equation by variation of parameters: $y'' + y = \sin^2(x)$. Find $y(x) =$

Neetesh Kumar | October 27, 2024

Differential Equation Homework Help

This is the solution to Math 2A, section 13Z, Fall 2023 | WebAssign
Math002ACh4Sec06 (Homework) Question - 2
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Step-by-step solution:

To solve the differential equation $y'' + y = \sin^2(x)$ using a variation of parameters, we will proceed as follows:

Step 1: Solve the Homogeneous Equation

The associated homogeneous equation is: $y'' + y = 0$

The characteristic equation for this differential equation is: $r^2 + 1 = 0$

Solving for $r$: $r = \pm i$

Thus, the solutions to the homogeneous equation are: $y_h(x) = C_1 \cos(x) + C_2 \sin(x)$

Step 2: Rewrite the Non-Homogeneous Term

We are given a non-homogeneous term $\sin^2(x)$ on the right side. To apply variation of parameters, we first rewrite $\sin^2(x)$ using the identity:

$\sin^2(x) = \frac{1 - \cos(2x)}{2}$

Thus, the equation becomes:

$y'' + y = \frac{1}{2} - \frac{1}{2} \cos(2x)$

This allows us to split the problem into solving two separate equations. We write the particular solution as:

$y_p(x) = y_{p1}(x) + y_{p2}(x)$

where:

1. $y_{p1}'' + y_{p1} = \frac{1}{2}$
2. $y_{p2}'' + y_{p2} = -\frac{1}{2} \cos(2x)$

Step 3: Solve for $y_{p1}(x)$

For the first part, we solve $y_{p1}'' + y_{p1} = \frac{1}{2}$.

Since the right side is a constant, we can try a constant solution, $y_{p1} = A$, where $A$ is a constant. Substitute into the equation:

$0 + A = \frac{1}{2}$

Thus, $A = \frac{1}{2}$, so:

$y_{p1}(x) = \frac{1}{2}$

Step 4: Solve for $y_{p2}(x)$ using Variation of Parameters

Now we solve $y_{p2}'' + y_{p2} = -\frac{1}{2} \cos(2x)$.

Since the solutions to the homogeneous equation are $\cos(x)$ and $\sin(x)$, we assume a particular solution of the form:

$y_{p2}(x) = u_1(x) \cos(x) + u_2(x) \sin(x)$

where $u_1(x)$ and $u_2(x)$ are functions to be determined.

For variation of parameters, we impose the conditions:

1. $u_1'(x) \cos(x) + u_2'(x) \sin(x) = 0$
2. $-u_1'(x) \sin(x) + u_2'(x) \cos(x) = -\frac{1}{2} \cos(2x)$

Step 5: Set Up and Solve for $u_1'(x)$ and $u_2'(x)$

The Wronskian $W(\cos(x), \sin(x))$ is: $W = \cos^2(x) + \sin^2(x) = 1$

Using Cramer’s rule, we find $u_1'(x)$ and $u_2'(x)$:

1. $u_1'(x) = \frac{\begin{vmatrix} 0 & \sin(x) \\ -\frac{\cos(2x)}{2} & \cos(x) \end{vmatrix}}{W} = \frac{\sin(x) \cos(2x)}{2}$

2. $u_1'(x) = \frac{\begin{vmatrix} \cos(x) & 0 \\ -\sin(x) & -\frac{\cos(2x)}{2} \end{vmatrix}}{W} = -\frac{\cos(x) \cos(2x)}{2}$

Step 6: Integrate to Find $u_1(x)$ and $u_2(x)$

Integrate $u_1'(x)$ and $u_2'(x)$ with respect to $x$:

1. $u_1(x) = \int \frac{\sin(x) \cos(2x)}{2} \, dx$

   So, $u_1(x) = \frac{3\cos(x)-\cos(3x)}{12}$.

2. $u_2(x) = \int -\frac{\cos(x) \cos(2x)}{2} \, dx$

   So, $u_2(x) = \frac{3\sin(x)+\sin(3x)}{12}$.

Step 7: Form the Particular Solution $y_p(x)$

$y_p(x) = y_{p1}(x) + y_{p2}(x) = \frac{1}{2} + y_{p2}(x)$

Substitute $u_1(x)$ and $u_2(x)$ into $y_{p2}(x) = u_1(x) \cos(x) + u_2(x) \sin(x)$:

$y_{p2}(x) = \left(\frac{3\cos(x)-\cos(3x)}{12}\right) \cos(x) + \left(\frac{3\sin(x)+\sin(3x)}{12}\right) \sin(x)$

Simplify each term:

$y_{p2}(x) = \frac{1}{4} - \frac{\cos(2x)}{12}$

so, $y_p(x) = \frac{1}{2} + \frac{1}{4} - \frac{\cos(2x)}{12} = \frac{3}{4} - \frac{\cos(2x)}{12}$

Step 8: Write the General Solution

The general solution to the differential equation is:

$y(x) = y_h(x) + y_p(x)$

where

$y_h(x) = C_1 \cos(x) + C_2 \sin(x)$

and

$y_p(x) = \frac{3}{4} - \frac{\cos(2x)}{12}$

So,

$y(x) = \boxed{C_1 \cos(x) + C_2 \sin(x) + \frac{3}{4} - \frac{\cos(2x)}{12}}$

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