Chapter 6, Problem 60RE

To determine

To prove: The statement: $y={\displaystyle {\int }_0^x\sin }\left(t^2\right)dt+x^3+x+2$

Explanation of Solution

Given information:

The differential equation is:

  $y^{\prime \prime }=2x\cos \left(x^2\right)+6x$

And the initial condition:

  $\begin{array}{c}{y}^{\prime }(0)=1,\\ y(0)=2\end{array}$

Calculation:

Separable the variables and anti differentiate as follows.

  $\begin{array}{c}{y}^{\prime }={\displaystyle \int \left(2x\cos \left(x^2\right)+6x\right)}dx\\ \to y^{\prime }={\displaystyle \int \cos }udu+{\displaystyle \int 6}xdx\\ \to y^{\prime }=\sin u+6\cdot \frac{x^{1+1}}{1+1}+C\\ \to y^{\prime }=\sin \left(x^2\right)+3x^2+C\end{array}$

Then apply the initial condition to find $C$:

Put,

  $\begin{array}{c}{y}^{\prime }=1\\ x=0\end{array}$

Then simplify,

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

Separable the variables and anti differentiate as follows.

  $\begin{array}{c}y={{\displaystyle \int }}^{\text{}}\left(\sin \left(x^2\right)+3x^2+1\right)dx\\ \to y={{\displaystyle \int }}^{\text{}}\sin \left(x^2\right)dx+3{{\displaystyle \int }}^{\text{}}{x}^{2}dx+{{\displaystyle \int }}^{\text{}}1dx\\ \to y={{\displaystyle \int }}^{\text{}}\sin \left(x^2\right)dx+3\cdot \frac{x^{2+1}}{2+1}+x+C\\ \to y={{\displaystyle \int }}^{\text{}}\sin \left(x^2\right)dx+x^3+x+C\end{array}$

Then apply the initial condition to find $C$:

Put,

  $\begin{array}{c}y=2\\ x=0\end{array}$

Then simplify,

  $\begin{array}{c}2={{\displaystyle \int }}^{\text{}}\sin \left(0^2\right)dx+0^3+0+C\\ \to 2=0+0+0\\ \to C=2\end{array}$

Use the Fundamental Theorem of Calculus to obtain $y={\displaystyle {\int }_0^x\sin }\left(t^2\right)dt+x^3+x+2$ ,

Then differentiate again and also verify the initial conditions.