Solve the differential equation by variation of parameters: $y'' + y = \sin(x)$. 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 - 1
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Step-by-step solution:

To solve the differential equation $y'' + y = \sin(x)$ using variation of parameters, we will proceed with the following steps.

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: Set Up Variation of Parameters

We seek a particular solution of the form: $y_p(x) = u_1(x) \cos(x) + u_2(x) \sin(x)$

where $u_1(x)$ and $u_2(x)$ are functions to be determined. For variation of parameters, we impose the conditions: $u_1'(x) \cos(x) + u_2'(x) \sin(x) = 0$ $-u_1'(x) \sin(x) + u_2'(x) \cos(x) = \sin(x)$

Step 3: Solve for $u_1'(x)$ and $u_2'(x)$

From the second equation, we can solve for $u_1'(x)$ and $u_2'(x)$ using Cramer's rule.

The Wronskian $W(\cos(x), \sin(x))$ is: $W = \begin{vmatrix} \cos(x) & \sin(x) \\ -\sin(x) & \cos(x) \end{vmatrix} = \cos^2(x) + \sin^2(x) = 1$

Now we can solve for $u_1'(x)$ and $u_2'(x)$:

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

   So, $u_1'(x) = -\sin^2(x)$.

2. $u_2'(x) = \frac{\begin{vmatrix} \cos(x) & 0 \\ -\sin(x) & \sin(x) \end{vmatrix}}{W} = \frac{\cos(x) \cdot \sin(x) - 0}{1} = \cos(x) \sin(x)$

   So, $u_2'(x) = \cos(x) \sin(x)$.

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

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

1. $u_1(x) = \int -\sin^2(x) \, dx$

   Using the identity $\sin^2(x) = \frac{1 - \cos(2x)}{2}$:

   $u_1(x) = -\int \frac{1 - \cos(2x)}{2} \, dx = -\frac{x}{2} + \frac{\sin(2x)}{4}$

   So, $u_1(x) = -\frac{x}{2} + \frac{\sin(2x)}{4}$.

2. $u_2(x) = \int \cos(x) \sin(x) \, dx$

   Using the substitution $u = \cos(x)$, $du = -\sin(x) \, dx$:

   $u_2(x) = -\int u \, du = -\frac{u^2}{2} = -\frac{\cos^2(x)}{2}$

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

Step 5: Form the Particular Solution $y_p(x)$

Substitute $u_1(x)$ and $u_2(x)$ into $y_p(x) = u_1(x) \cos(x) + u_2(x) \sin(x)$:

$y_p(x) = \left(-\frac{x}{2} + \frac{\sin(2x)}{4}\right) \cos(x) - \frac{\cos^2(x)}{2} \sin(x)$

Simplify each term:

1. $\left(-\frac{x}{2}\right) \cos(x) = -\frac{x}{2} \cos(x)$
2. $\frac{\sin(2x)}{4} \cos(x) = \frac{1}{4} \sin(2x) \cos(x)$
3. $\frac{\cos^2(x)}{2} \sin(x) = \frac{\cos^2(x)\sin(x)}{2}$

Thus, the particular solution is:

$y_p(x) = -\frac{x}{2} \cos(x) + \frac{1}{4} \sin(2x) \cos(x) - \frac{\cos^2(x)\sin(x)}{2}$

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

Step 6: 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

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

So,

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

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