# 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

Become a Study.com member to unlock this answer! Create your account

View this answerGiven 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:\:...

See full answer below.

#### Ask a question

Our experts can answer your tough homework and study questions.

Ask a question Ask a question#### Search Answers

#### Learn more about this topic:

from

Chapter 16 / Lesson 3In 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.