# Solve the initial value problem. y'' - 2y' + 10y equals 0 ; y(0) equals -1 , y'(0) equals 5

## Question:

Solve the initial value problem.

{eq}y'' - 2y' + 10y = 0 ; y(0) = -1 , y'(0) = 5 {/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 solution of the differential equation in the form as follows

{eq}y(x)=e^{\alpha x}\left [ c_1\cos (\beta x) +c_2\sin (\beta x)\right ] {/eq}

Now, we apply initial conditions we get required solution.

Consider the differential equation

{eq}y'' - 2y' + 10y = 0 ;\quad y(0) = -1 ,\: \: y'(0) = 5 {/eq}

Rewrite the differential equation as follows

{eq}(D^{2}-2D+10)y=0 {/eq}

The auxiliary equation corresponding to homogeneous differential equation is as follows

{eq}m^{2}-2m+10=0\\ \Rightarrow m=1\pm 3i {/eq}

Therefore, the solution corresponding to homogeneous differential equation is as follows

{eq}y(x)=e^{x}(c_1\cos x +c_2\sin x)\\ y{}'(x)=e^x\left(c_1\cos \left(x\right)+c_2\sin \left(x\right)\right)+\left(-c_1\sin \left(x\right)+c_2\cos \left(x\right)\right)e^x\\ y(0) = -1\Rightarrow c_1=-1\\ y'(0) = 5\Rightarrow c_1+c_2=5\Rightarrow c_2=6 {/eq}

Hence,

{eq}y(x)=e^{x}(-\cos x +6\sin x) {/eq}

First-Order Linear Differential Equations

from

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.