# Find the solution of the differential equation: y double prime + y = sin x.

## Question:

Find the solution of the differential equation: {eq}{y}'' + y = \sin x {/eq}.

## Solving Second Order Differential Equation:

A second-order differential equation of the form {eq}y''+ay'+by=g(x) {/eq} can also be expressed as {eq}(D^2+aD+b)y=g(x) {/eq}, where {eq}D {/eq} represents the first derivative and {eq}D^2 {/eq} represents the second derivative.

So our differential equation becomes {eq}f(D)y=g(x) {/eq}.

Now the general solution is given by {eq}y(x)= y_h+y_p {/eq}.

{eq}y_h {/eq} is the solution of the homogeneous differential equation {eq}f(D)=0 {/eq}.

{eq}y_p {/eq} is the particular integral or particular solution.

If {eq}g(x)=\sin ax {/eq}, then particulat solution is given by:

{eq}y_p=\left\{\begin{matrix} \frac{1}{f(-a^2)} \sin ax & ,\text{if}\, \,f(-a^2)\neq 0\\ \frac{-x}{2a} \cos ax & ,\text{if}\, \,f(-a^2)= 0 \end{matrix}\right. {/eq}

That is we need to substitute {eq}-a^2 {/eq} in place of {eq}D^2 {/eq} in {eq}f(D^2) {/eq}.

Note: In the case where {eq}g(x) {/eq} is a {eq}sine {/eq} function, we put {eq}f(D^2) {/eq} instead of {eq}f(D) {/eq}, if differential expression contins {eq}D^2 {/eq} terms only. Otherwise we put {eq}f(D^2,D) {/eq}

We need to find the general solution of {eq}{y}'' + y = \sin x {/eq}.

This equation can be rewritten as {eq}D^2y+y=\sin x \Rightarrow (D^2+1)y=\sin...

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