Chapter 3.1, Problem 1P

In Problems 1 through 16, a homogeneous second-order linear differential equation, two functions $y_1\text{and}{y}_2$, and a pair of initial conditions are given. First verify that $y_1\text{and}{y}_2$ are solutions of the differential equation. Then find a particular solution of the form $y=c_1{y}_1+c_2{y}_2$ that satisfies the given initial conditions. Prunes denote derivatives with respect to x.

$y"-y=0;y_1=e^x,y_2=e^{-x};y\left(0\right)=0,y\text{'}\left(0\right)=5$

Program Plan Intro

Program Description: Purpose of problem is to verify that $y_1$ and $y_2$ are solutions of the differential equation $y^{″}-y=0$ and also find a particular solution of differential equation in the form of $y=c_1{y}_1+c_2{y}_2$ .

Explanation of Solution

Given information:

The homogeneous second order differential equation is $y^{″}-y=0$ .

The value of $y_1$ is $e^x$ and $y_2$ is $e^{-x}$ .

The initial condition is $y\left(0\right)=0$ and $y^{\prime }\left(0\right)=5$ .

Explanation:

The given differential equation can be represented as,

  $\frac{d^{2}y}{d{x}^2}-y=0$ ....... (1)

Substitute $e^x$ for $y$ in equation (1),

  $\begin{array}{l}\frac{d^2\left(e^x\right)}{d{x}^2}-e^x\stackrel{?}{=}0\\ \frac{d\left(e^x\right)}{dx}-e^x\stackrel{?}{=}0\\ e^x-e^x\stackrel{?}{=}0\\ 0=0\end{array}$

Therefore, it is verified that $y_1$ is the solution of the differential equation $y^{″}-y=0$ .

Substitute $e^{-x}$ for $y$ in equation (1),

  $\begin{array}{l}\frac{d^2\left(e^{-x}\right)}{d{x}^2}-e^{-x}\stackrel{?}{=}0\\ -\frac{d\left(e^{-x}\right)}{dx}-e^{-x}\stackrel{?}{=}0\\ e^{-x}-e^{-x}\stackrel{?}{=}0\\ 0=0\end{array}$

Therefore, it is verified that $y_2$ is the solution of the differential equation $y^{″}-y=0$ .

The solution of the differential equation can be written as,

  $y=c_1{y}_1+c_2{y}_2$ ....... (2)

Substitute, $e^x$ for $y_1$ and $e^{-x}$ for $y_2$ in equation (2),

  $y=c_1{e}^x+c_2{e}^{-x}$ ....... (3)

Differentiate equation (3) with respect to x as shown below.

  $y^{\prime }=c_1{e}^x-c_2{e}^{-x}$ ....... (4)

Apply first initial condition $y\left(0\right)=0$ in equation (3).

  $\begin{array}{c}0=c_1{e}^0+c_2{e}^{-0}\\ 0=c_1+c_2\\ c_1=-c_2\end{array}$

Apply second initial condition $y^{\prime }\left(0\right)=5$ in equation (4).

  $\begin{array}{c}5=c_1{e}^0-c_2{e}^{-0}\\ 5=c_1-c_2\end{array}$

Substitute $\left(-c_2\right)$ for $c_1$ in above equation.

  $\begin{array}{l}5=-c_2-c_2\\ 5=-2c_2\\ c_2=-\frac{5}{2}\end{array}$

Therefore, the value of $c_1$ can be obtained as,

  $\begin{array}{c}{c}_1=-c_2\\ =-\left(-\frac{5}{2}\right)\\ =\frac{5}{2}\end{array}$

Substitute $\frac{5}{2}$ for $c_1$ and $-\frac{5}{2}$ for $c_2$ in equation (3),

  $y=\frac{5}{2}{e}^x-\frac{5}{2}{e}^{-x}$

Conclusion:

Thus, the solution of differential equation $y^{″}-y=0$ is $y=\frac{5}{2}{e}^x-\frac{5}{2}{e}^{-x}$ .