# Solve the differential equation y'' - 4y' +13 = e^{2x} \cos 3x subject to the initial...

## Question:

Solve the differential equation {eq}y'' - 4y' +13 = e^{2x} \cos 3x {/eq} subject to the initial conditions {eq}y(0) = 0 \enspace and \enspace y'(0) = 1 {/eq}

## Linear Initial Value Problem:

To solve an IVP problem like a second order linear problem, we first need to find its characteristic equation (polynomial) by replacing {eq}y^{(n)} \text { with } r^n {/eq} and solving the characteristic polynomial equation for {eq}r {/eq}:

- If the solution is a real number, then the solution to the differential equation is {eq}y=Ce^{rt} {/eq}.

- If the solution is a real double root, then the solution to the differential equation is {eq}y=C_1e^{rt}+t\,C_2e^{rt} {/eq}.

- If the solution is not a real number, {eq}r=\alpha+i\beta {/eq}, then the solution to the differential equation is {eq}y=C_1e^{\alpha t}\cos(\beta t)+\,C_2e^{\alpha t}\sin(\beta t) {/eq}.

For a non-homogeneous equation such as {eq}y'' - 4y' +13 = e^{2x} \cos 3x\\ y'' - 4y' =- 13 +e^{2x} \cos 3x, {/eq} we need to find its complementary solution {eq}y_c(x). {/eq} The complementary solution is a solution to homogeneous equation {eq}y'' - 4y' = 0. {/eq}

Then we need to find its particular solution {eq}y'' - 4y' = g(x). {/eq} Since {eq}g(x)=e^{2x} \cos 3x -13 {/eq} then its particular solution is {eq}y_p(x)=(Dx+E)\;+\;e^{2x} (A\cos 3x+B\sin 3x). {/eq}

The general solution can be obtained by {eq}y(x)=y_p(x)+y_c(x) {/eq}

## Answer and Explanation: 1

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View this answerFind the complementary solution {eq}y'' - 4y' = 0. {/eq} Its characteristic equation is {eq}r^2-4r= 0\\ r(r-4)=0. {/eq}

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Chapter 11 / Lesson 13Learn to define the initial value problem and initial value formula. Learn how to solve initial value problems in calculus. See examples of initial value problems.