Chapter 7.1, Problem 5E

a.

To determine

To identify : Whether the given function is the solution of the differential equation $y^{″}+y=\sin x$ or not.

Answer to Problem 5E

  $y=\sin x$ is not a solution of differential equation (i).

Explanation of Solution

Given information :

  $y^{″}+y=\sin x\mathrm{........}\left(i\right)$

The given function is $y=\sin x$

Formula used :

If $y=f\left(x\right)$ is a solution of a differential equation then it must satisfy the equation.

  $\frac{d}{dx}\left(\sin x\right)=\cos x$ ; $\frac{d}{dx}\left(\cos x\right)=-\sin x$

Calculation :

  $\begin{array}{l}y=\sin x\\ y^{\prime }=\cos x\\ y^{″}=-sinx\end{array}$

Therefore, $y^{″}+y=-\sin x+\sin x=0\ne \sin x$

Hence, $y=\sin x$ is not a solution of differential equation (i)

b.

To determine

To identify : Whether the given function is the solution of the differential equation $y^{″}+y=\sin x$ or not.

Answer to Problem 5E

  $y=\cos x$ is not a solution of differential equation (i)

Explanation of Solution

Given information :

  $y^{″}+y=\sin x\mathrm{........}\left(i\right)$

The given function is $y=\cos x$

Formula used :

If $y=f\left(x\right)$ is a solution of a differential equation then it must satisfy the equation.

  $\frac{d}{dx}\left(\sin x\right)=\cos x$ ; $\frac{d}{dx}\left(\cos x\right)=-\sin x$

Calculation :

  $\begin{array}{l}y=\cos x\\ y^{\prime }=-\sin x\\ y^{″}=-\cos x\end{array}$

Therefore, $y^{″}+y=-\cos x+\cos x=0\ne \sin x$

Hence, $y=\cos x$ is not a solution of differential equation (i)

c.

To determine

To identify : Whether the given function is the solution of the differential equation $y^{″}+y=\sin x$ or not.

Answer to Problem 5E

  $y=\frac{1}{2}x\sin x$ is not a solution of differential equation (i)

Explanation of Solution

Given information :

  $y^{″}+y=\sin x\mathrm{........}\left(i\right)$

The given function is $y=\frac{1}{2}x\sin x$

Formula used :

If $y=f\left(x\right)$ is a solution of a differential equation then it must satisfy the equation.

  $\frac{d}{dx}\left(\sin x\right)=\cos x$ ; $\frac{d}{dx}\left(\cos x\right)=-\sin x$

  $\frac{d}{dx}\left(uv\right)=u\frac{d}{dx}v+v\frac{d}{dx}u$

Calculation :

  $\begin{array}{l}y=\frac{1}{2}x\sin x\\ \Rightarrow y^{\prime }=\frac{1}{2}\sin x+\frac{1}{2}x\cos x\\ \Rightarrow y^{″}=\frac{1}{2}\cos x+\frac{1}{2}\cos x-\frac{1}{2}x\sin x\\ \Rightarrow y^{″}=\cos x-\frac{1}{2}x\sin x\end{array}$

Therefore, $y^{″}+y=\cos x-\frac{1}{2}x\sin x+\frac{1}{2}x\sin x=\cos x\ne \sin x$

Hence, $y=\frac{1}{2}x\sin x$ is not a solution of differential equation (i)

d.

To determine

To identify : Whether the given function is the solution of the differential equation $y^{″}+y=\sin x$ or not.

Answer to Problem 5E

  $y=-\frac{1}{2}x\cos x$ is a solution of differential equation (i)

Explanation of Solution

Given information :

  $y^{″}+y=\sin x\mathrm{........}\left(i\right)$

The given function is $y=-\frac{1}{2}x\cos x$

Formula used :

If $y=f\left(x\right)$ is a solution of a differential equation then it must satisfy the equation.

  $\frac{d}{dx}\left(\sin x\right)=\cos x$ ; $\frac{d}{dx}\left(\cos x\right)=-\sin x$

  $\frac{d}{dx}\left(uv\right)=u\frac{d}{dx}v+v\frac{d}{dx}u$

Calculation :

  $\begin{array}{l}y=-\frac{1}{2}x\cos x\\ \Rightarrow y^{\prime }=-\frac{1}{2}\cos x+\frac{1}{2}x\sin x\\ \Rightarrow y^{″}=\frac{1}{2}\sin x+\frac{1}{2}\sin x+\frac{1}{2}x\cos x\end{array}$

Therefore, $y^{″}+y=\frac{1}{2}\sin x+\frac{1}{2}\sin x+\frac{1}{2}x\sin x-\frac{1}{2}x\sin x=\sin x$

Hence, $y=-\frac{1}{2}x\cos x$ is a solution of differential equation (i)