Given the equation:
$$x y^{3}{\left(x \right)} + e^{x^{2}} \frac{d}{d x} y{\left(x \right)} = 0$$
This differential equation has the form:
f1(x)*g1(y)*y' = f2(x)*g2(y),
where
$$\operatorname{f_{1}}{\left(x \right)} = 1$$
$$\operatorname{g_{1}}{\left(y \right)} = 1$$
$$\operatorname{f_{2}}{\left(x \right)} = - x e^{- x^{2}}$$
$$\operatorname{g_{2}}{\left(y \right)} = y^{3}{\left(x \right)}$$
We give the equation to the form:
g1(y)/g2(y)*y'= f2(x)/f1(x).
Divide both parts of the equation by g2(y)
$$y^{3}{\left(x \right)}$$
we get
$$\frac{\frac{d}{d x} y{\left(x \right)}}{y^{3}{\left(x \right)}} = - x e^{- x^{2}}$$
We separated the variables x and y.
Now, multiply the both equation sides by dx,
then the equation will be the
$$\frac{dx \frac{d}{d x} y{\left(x \right)}}{y^{3}{\left(x \right)}} = - dx x e^{- x^{2}}$$
or
$$\frac{dy}{y^{3}{\left(x \right)}} = - dx x e^{- x^{2}}$$
Take the integrals from the both equation sides:
- the integral of the left side by y,
- the integral of the right side by x.
$$\int \frac{1}{y^{3}}\, dy = \int \left(- x e^{- x^{2}}\right)\, dx$$
Detailed solution of the integral with yDetailed solution of the integral with xTake this integrals
$$- \frac{1}{2 y^{2}} = Const + \frac{e^{- x^{2}}}{2}$$
Detailed solution of the equationWe get the simple equation with the unknown variable y.
(Const - it is a constant)
Solution is:
$$\operatorname{y_{1}} = y{\left(x \right)} = - \sqrt{- \frac{e^{x^{2}}}{C_{1} e^{x^{2}} + 1}}$$
$$\operatorname{y_{2}} = y{\left(x \right)} = \sqrt{- \frac{e^{x^{2}}}{C_{1} e^{x^{2}} + 1}}$$