91Ó°ÊÓ

In mathematics, a differential equation is an equation that relates one or further unknown functions and their derivations. In operations, the functions generally represent physical amounts, the derivations represent their rates of change, and the differential equation defines a relationship between the two.

The given differential equation is$dy+\left(2xy-x{e}^{-x^2}\right)dx=0$.

There need to find the general solution of the given differential equation and compare the solution with computer solution.

Firstly, there need to write this differential equation to make it in the form$y^{\text{'}}+Py=Q$, that is

$$\begin{gathered} \frac{dy}{dx}=x{e}^{-x^2}-2xy \\ {y}^{\text{'}}+2xy=x{e}^{-x^2} \end{gathered}$$

According to the above differential equation $P\left(x\right)=2x$and$Q\left(x\right)=x{e}^{-x^2}.$

Now there need to find I, then the integration factor,

$$\begin{gathered} I=∫Pdx \\ =∫2xdx \\ =x^2 \end{gathered}$$

Further, the integrating factor is

$e^I=e^{x^2}$

Substituting the values in the formula$y{e}^I=∫Q{e}^{I}dx+c$

Now, the solution of the given differential equation is

$$\begin{gathered} y{e}^I=∫Q{e}^{I}dx \\ =∫\left(x{e}^{-x^2}\right)e^{x^2}dx \\ =\frac{x^2}{2}+C \end{gathered}$$

So, the general solution is$y=\frac{1}{2}{x}^2{e}^{-x^2}+C{e}^{-x^2}$

Further, by using the wolfram Mathematica the general solution of the given differential equation is

On comparing the solution finding by the wolfram Mathematica and the solution finding above both will give the same results.

Thus, the general solution of the given differential equation$dy+\left(2xy-x{e}^{-x^2}\right)dx=0$is $y=\frac{1}{2}{x}^2{e}^{-x^2}+C{e}^{-x^2}$