Chapter 7, Problem 60RE

To determine

To Show:

To show that $y={\displaystyle \underset{0}{\overset{x}{\int }}\sin \left(t^2\right)}dt+x^3+x+2$ is the solution of the initial value problem $y^{″}=2x\cos \left(x^2\right)+6x$ with initial conditions $y^{\prime }\left(0\right)=1;y\left(0\right)=2$ .

Explanation of Solution

Given information:

The differential equation is $y^{″}=2x\cos \left(x^2\right)+6x$ with initial conditions $y^{\prime }\left(0\right)=1;y\left(0\right)=2$ .

The given differential equation is

  $y^{″}=2x\cos \left(x^2\right)+6x$

Integrating with respect to x,

  $y^{\prime }={\displaystyle \int 2x\cos \left(x^2\right)dx+3x^2+C}$

Substitute $u=x^2\Rightarrow du=2xdx$ ,

  $\begin{array}{l}{y}^{\prime }={\displaystyle \int \cos udu+3x^2+C}\\ y^{\prime }=\sin \left(x^2\right)+3x^2+C\end{array}$

Now substitute the initial conditions $y^{\prime }\left(0\right)=1$ ;

  $\begin{array}{l}{y}^{\prime }=\sin \left(x^2\right)+3x^2+C\\ 1=0+0+C\\ C=1\end{array}$

  $y^{\prime }=\sin \left(x^2\right)+3x^2+1$

Integrating with respect to x,

  $y={\displaystyle \int \sin \left(x^2\right)dx}+x^3+x+C$

Now substitute the initial conditions $y\left(0\right)=2$ ;

  $\begin{array}{l}y={\displaystyle \int \sin \left(x^2\right)dx}+x^3+x+C\\ 2=0+0+0+C\\ C=2\end{array}$

Now taking limit from $0$ to $x$ ,

  $y={\displaystyle \underset{0}{\overset{x}{\int }}\sin \left(t^2\right)dt+x^3+x+2}$

Hence the proof.