Question

In Problems 27–30 use (12) of Section 1.1 to verify that the indicatedfunction is a solution of the given differential equation. Assume anappropriate interval I of definition of each solution.

$\mathit{y}\mathbf{+}\mathit{y}\mathbf{=}{\mathit{e}}^{{\mathbf{x}}^{\mathbf{2}}}\mathbf{;}\mathit{y}\mathbf{=}\mathit{s}\mathit{i}\mathit{n}\mathit{x}\underset{\mathbf{0}}{\overset{\mathbf{x}}{\mathbf{\int }}}{\mathit{e}}^{{\mathbf{t}}^{\mathbf{2}}}\mathit{c}\mathit{o}\mathit{s}\mathit{t}\mathit{d}\mathit{t}\mathbf{-}\mathit{c}\mathit{o}\mathit{s}\mathit{x}\underset{\mathbf{0}}{\overset{\mathbf{x}}{\mathbf{\int }}}{\mathit{e}}^{{\mathbf{t}}^{\mathbf{2}}}\mathit{s}\mathit{i}\mathit{n}\mathit{t}\mathit{d}\mathit{t}$

Short answer

The function $y=\sin x\underset{0}{\overset{x}{\int }}{e}^{t^2}\cos tdt-\cos x\underset{0}{\overset{x}{\int }}{e}^{t^2}\sin tdt$ is the solution for the given differential equation.

Step-by-step solution

Definition of differential equation.

An equation containing the derivatives of one or more unknown functions (or dependent variables), with respect to one or more independent variables, is said to be a differential equation (DE).

Take differential of both sides of the given equations.

Take differential of both sides of the given equations:

$\begin{array}{rcl}\frac{d}{dx}\left(y\right)& =& \frac{d}{dx}\left(\sin x\underset{0}{\overset{x}{\int }}{e}^{t^2}\cos tdt-\cos x\underset{0}{\overset{x}{\int }}{e}^{t^2}\sin tdt\right)\\ y\text{'}& =& \cos x\underset{0}{\overset{x}{\int }}{e}^{t^2}\cos tdt+\sin x{e}^{x^2}\cos x+\sin x\underset{0}{\overset{x}{\int }}{e}^{t^2}\sin tdt-\cos x{e}^{x^2}\sin x\\ y\text{'}& =& \cos x\underset{0}{\overset{x}{\int }}{e}^{t^2}\cos tdt+\sin x\underset{0}{\overset{x}{\int }}{e}^{t^2}\sin tdt\end{array}$

Take differential of both sides.

$\begin{array}{rcl}y& =& -\sin x\underset{0}{\overset{x}{\int }}{e}^{t^2}\cos tdt+\cos x{e}^{x^2}\cos x+\cos x\underset{0}{\overset{x}{\int }}{e}^{t^2}\sin tdt+\sin x{e}^{x^2}\sin x\\ y& =& -\sin x\underset{0}{\overset{x}{\int }}{e}^{t^2}\cos tdt+e^{x^2}({\cos }^{2}x+{\sin }^{2}x)+\cos x\underset{0}{\overset{x}{\int }}{e}^{t^2}\sin tdt\\ y& =& -\sin x\underset{0}{\overset{x}{\int }}{e}^{t^2}\cos tdt+\cos x\underset{0}{\overset{x}{\int }}{e}^{t^2}\sin tdt+e^{x^2}..........\left(1\right)\end{array}$

Adding (1) and $y=\sin x\underset{0}{\overset{x}{\int }}{e}^{t^2}\cos tdt-\cos x\underset{0}{\overset{x}{\int }}{e}^{t^2}\sin tdt$

So, $y+y=-\sin x\underset{0}{\overset{x}{\int }}{e}^{t^2}\cos tdt+\cos x\underset{0}{\overset{x}{\int }}{e}^{t^2}\sin tdt+e^{x^2}+\sin x\underset{0}{\overset{x}{\int }}{e}^{t^2}\cos tdt-\cos x\underset{0}{\overset{x}{\int }}{e}^{t^2}\sin tdt$

Simplify we get,

$y+y=e^{x^2}$

Therefore, the indicated function is the solution for the given differential equation.

So, $y=\sin x\underset{0}{\overset{x}{\int }}{e}^{t^2}\cos tdt-\cos x\underset{0}{\overset{x}{\int }}{e}^{t^2}\sin tdt$is the solution for the given differential equation.