# Solve. \\ y''+y'+\frac{y}{4}=1+xe^x when y(0)=0, \ \ y'(0)=1

## Question:

Solve.

{eq}y''+y'+\frac{y}{4}=1+xe^x{/eq} when {eq}y(0)=0, \ \ y'(0)=1{/eq}

## Differential Equation:

Differential equation of form {eq}\displaystyle \phi \left(D\right)y=f\left(x\right)\:where,\:D=\frac{d}{dx} {/eq} has solution {eq}\displaystyle y=y_c+y_p {/eq}

Where, complimentary function {eq}\displaystyle y_c {/eq} is depend on auxiliary equation and

particular integral {eq}\displaystyle y_p {/eq} is depend on function f(x).

Use the initial conditions to find the value of constants.

## Answer and Explanation: 1

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Given ODE is

{eq}\displaystyle y''+y'+\frac{y}{4}=1+xe^x\\ \displaystyle \left(D^2+D+\frac{1}{4}\right)y=1+xe^x\\ \displaystyle A.E:\:...

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#### Learn more about this topic: First-Order Linear Differential Equations

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Chapter 16 / Lesson 3
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In this lesson you'll learn how to solve a first-order linear differential equation. We first define what such an equation is, and then we give the algorithm for solving one of that form. Specific examples follow the more general description of the method.