Question

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

$\frac{\mathbf{d}\mathbf{y}}{\mathbf{d}\mathbf{x}}\mathbf{-}\mathbf{2}\mathit{x}\mathit{y}\mathbf{=}{\mathit{e}}^{\mathbf{x}}\mathbf{;}\mathit{y}\mathbf{=}{\mathit{e}}^{{\mathbf{x}}^{\mathbf{2}}}{\mathbf{\int }}_{\mathbf{0}}^{\mathbf{x}}{\mathit{e}}^{\mathbf{t}\mathbf{-}{\mathbf{t}}^{\mathbf{2}}}\mathit{d}\mathit{t}$

Short answer

The indicated function is a solution of the differential function.

Step-by-step solution

Simplify the given differential equation

Let the given differential equation be$\mathrm{y}={\mathrm{e}}^{{\mathrm{x}}^2}{\int }_0^{\mathrm{x}}{\mathrm{e}}^{\mathrm{t}-{\mathrm{t}}^2}\mathrm{dt}$.

Multiply each side of the equation by ${\mathrm{e}}^{-{\mathrm{x}}^2}$.

$\begin{array}{c}{\mathrm{ye}}^{-{\mathrm{x}}^2}={\mathrm{e}}^{-{\mathrm{x}}^2}{\mathrm{e}}^{{\mathrm{x}}^2}{\int }_0^{\mathrm{x}}{\mathrm{e}}^{\mathrm{t}-{\mathrm{t}}^2}\mathrm{dt}\\ {\mathrm{ye}}^{-{\mathrm{x}}^2}={\int }_0^{\mathrm{x}}{\mathrm{e}}^{\mathrm{t}-{\mathrm{t}}^2}\mathrm{dt}\end{array}$

Determine the solution of the indicated function

Take differential on both sides of the equation.

$\begin{array}{c}\frac{\mathrm{d}}{\mathrm{dx}}\left({\mathrm{ye}}^{-{\mathrm{x}}^2}\right)=\frac{\mathrm{d}}{\mathrm{dx}}{\int }_0^{\mathrm{x}}{\mathrm{e}}^{\mathrm{t}-{\mathrm{t}}^2}\mathrm{dt}\\ \frac{\mathrm{dy}}{\mathrm{dx}}\left({\mathrm{e}}^{-{\mathrm{x}}^2}\right)-2{\mathrm{xye}}^{-{\mathrm{x}}^2}={\mathrm{e}}^{\mathrm{x}-{\mathrm{x}}^2}\end{array}$

Multiply${\mathrm{e}}^{{\mathrm{x}}^2}$ on both sides of the equation.

$\frac{\mathrm{dy}}{\mathrm{dx}}-2\mathrm{xy}={\mathrm{e}}^{\mathrm{x}}$

Hence, the indicated function is a solution to the differential function.