# Solve the initial-value problem {y}''- 2{y}' = x + 2e^x, y(0) = 13, {y}'(0) = \frac{23}{4}.

## Question:

Solve the initial-value problem {eq}{y}''- 2{y}' = x + 2e^x, y(0) = 13{/eq}, {eq}{y}'(0) = \frac{23}{4}{/eq}.

## Solution of the Differential Equation:

The given differential equation is a second order linear differential equation. To find the solution of the differential equation first we find the auxiliary equation and then write the complementary solution in the form as follows

{eq}\displaystyle y_c=c_1e^{\alpha x}+c_2e^{\beta x} {/eq}

To find the particular solution we use the following property as follows

{eq}\displaystyle \left [ \frac{1}{f(D)}e^{ax}=\frac{1}{f(a)}e^{ax},\: \: f(a)\neq 0 \right ]\\ \displaystyle \left [ \frac{1}{f(D)}e^{ax}=\frac{x}{f{}'(D)}e^{ax},\: \: f(a= 0 \right ] {/eq}

Where {eq}a {/eq} is a constant and {eq}D\equiv \frac{\mathrm{d} }{\mathrm{d} x} {/eq}

Then the general solution of the differential equation will be in the form of

{eq}\displaystyle y(x)=y_c+y_p{/eq}

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

Consider the differential equation

{eq}\displaystyle {y}''- 2{y}' = x + 2e^x,\: \: y(0) = 13,\quad y{}'(0) = \frac{23}{4} {/eq}

Rewrite the...

First-Order Linear Differential Equations

from

Chapter 16 / Lesson 3
2K

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.