Question

In Problems 29 and 30 proceed as in Example 5 and find an explicit solution of the given initial value problem.

$\frac{\mathbf{d}\mathbf{y}}{\mathbf{d}\mathbf{x}}\mathbf{=}\mathit{y}{\mathit{e}}^{\mathbf{-}{\mathbf{x}}^{\mathbf{2}}}\mathbf{,}\mathbf{}\mathbf{}\mathbf{}\mathit{y}\mathbf{\left(}\mathbf{4}\mathbf{\right)}\mathbf{=}\mathbf{1}$

Short answer

The explicit solution is $y\left(x\right)=e^{\underset{4}{\overset{x}{\int }}{e}^{-t^{2}dt}}.$.

Step-by-step solution

Definition

A first-order differential equation of the form $\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{=}\mathit{g}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathit{h}\mathbf{\left(}\mathit{y}\mathbf{\right)}$is said to be separable or to have separable variables.

Solve integral

Consider the initial value problem$\frac{dy}{dx}=y{e}^{-x^2},y\left(4\right)=1$;

Consider,$\frac{dy}{dx}=y{e}^{-x^2}$

Separating the variables$\frac{1}{y}dy=e^{-x^2}dx$

The function$\text{g}\left(\text{x}\right)=e^{-x^2}$ is continuous on $(-\infty ,\infty )$, but its anti derivative is not an elementary function. Usingas dummy variable of integration,

$\begin{array}{rcl}{\int }_4^x\frac{dy}{y}& =& {\int }_4^x{e}^{-t^2}dt\\ \left[\ln y\right(t){]}_4^x& =& {\int }_4^x{e}^{-t^2}dt\end{array}$

Apply limits

Applying lower & upper limits we get;

$\ln y\left(x\right)-\ln y\left(4\right)={\int }_4^x{e}^{-t^2}dt$

Add In y(4) on both sides then to get,

$\begin{array}{rcl}\ln y\left(x\right)-\ln y\left(4\right)+\ln y\left(4\right)& =& \ln y\left(4\right)+{\int }_4^x{e}^{-t^2}dt\ln y\left(x\right)\\ & =& \ln y\left(4\right)+{\int }_4^x{e}^{-t^2}dt\ln y\left(x\right)\\ & =& \ln \left(1\right)+{\int }_4^x{e}^{-t^2}dt\left(\text{since y}\right(4)=1)\\ \ln y\left(x\right)& =& 0+{\int }_4^x{e}^{-t^2}dt\end{array}$

$y\left(x\right)=e^{\underset{4}{\overset{x}{\int }}{e}^{-t^2}dt}$

Therefore, the explicit solution of the given initial value problem is$y\left(x\right)=e^{\underset{4}{\overset{x}{\int }}{e}^{-t^2}dt}$