Question

Proceed as in Example 7 and express the solution of the given initial value problem in terms of erf (Problem 43) and erfc (Problem 44).

$\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{-}\mathbf{2}\mathbf{xy}\mathbf{=}\mathbf{-}\mathbf{1}\mathbf{,}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\sqrt{\mathbf{\pi }}\mathbf{/}\mathbf{2}$

Short answer

So, the solution of the given initial value is $y=\frac{\sqrt{p}}{2}{e}^{x^2}\left[erf\right(x)-erf(1\left)\right]+e^{x^2-1}$.

Step-by-step solution

Form of linear equation

The linear differential equation is of the form $\frac{\mathbf{dy}}{\mathbf{dx}}+\mathbf{Py}=\mathbf{Q}$, where P and Qare numeric constants or functions in x.

Evaluation

Linear differential equation of the first order

$y^{\text{'}}+P\left(x\right)y=f\left(x\right)$

The integrating factor is given by

$e^{\int P\left(x\right)dx}$

The error function is given by

role="math" localid="1664264803101" $\begin{array}{l}erf\left(x\right)=\frac{2}{\sqrt{p}}{\int }_0^x{e}^{-t^2}dt\\ {\int }_0^x{e}^{-t^2}dt=\frac{\sqrt{p}}{2}erf\left(x\right)\end{array}$

and the complementary error function is given by

$erfc\left(x\right)=\frac{2}{\sqrt{p}}{\int }_x^{\backslash \mathrm{currency}}{e}^{-t^2}dt$

Based on (1) and the given differential equation, we obtain

$P\left(x\right)=-2x\mathrm{and}f\left(x\right)=-1$

Find integrated factor

Use the formula (2), to get integrating factor

$\begin{array}{l}{e}^{\int P\left(x\right)dx}=e^{-2\int xdx}\\ =e^{-z^2}\\ y{e}^{-x^2}=-{\int }_0^x{e}^{-t^2}dt+c\\ \stackrel{3}{=}-\frac{\sqrt{p}}{2}erf\left(x\right)+c\end{array}$

Use initial condition

Using the initial condition to get $y\left(0\right)=\frac{\sqrt{\pi }}{2}$

$\begin{array}{l}y\left(0\right)=-\frac{\sqrt{p}}{2}erf\left(0\right)e^0+c{e}^0\left(e^0-1\right)\\ \frac{\sqrt{p}}{2}=-\frac{\sqrt{p}}{2}erf\left(0\right)+c\\ c=\frac{\sqrt{p}}{2}\\ \left(erf\right(0)=0)\end{array}$

Substitution

Substitute the value $c=\frac{\sqrt{\pi }}{2}$in the equation (5)

$\begin{array}{l}y=-\frac{\sqrt{p}}{2}erf\left(x\right)e^{x^2}+\frac{\sqrt{p}}{2}{e}^{x^2}\\ y=\frac{\sqrt{p}}{2}{e}^{x^2}[1-erf(x\left)\right]\\ y=\frac{\sqrt{p}}{2}{e}^{x^2}erfc\left(x\right)\end{array}$

So, the required solution is $y=\frac{\sqrt{p}}{2}{e}^{x^2}erfc\left(x\right)$.