Question

Solve dy/dx + yx = y2e^(x^2/2)sinx

Answer 1

$\frac{dy}{dx}+yx=y^2e^{\left(\frac{x^2}{2}\right)}\sin \left(x\right)$

First order Bernoulli Ordinary Differential Equation

A first order Bernoulli ODE has the form of $y'+p\left(x\right)y=q\left(x\right)y^n$

Substitute $\frac{dy}{dx}\mathrm{\:with\:}y'\:$

$y'\:+yx=y^2e^{\left(\frac{x^2}{2}\right)}\sin \left(x\right)$

Rewrite in the form of a first order Bernoulli ODE:  $y'+p\left(x\right)y=q\left(x\right)y^n$

$p\left(x\right)=x,\:\quad q\left(x\right)=e^{\frac{x^2}{2}}\sin \left(x\right),\:\quad n=2$

The general solution is obtained by substituting $v=y^{1-n}$

and solving $\frac{1}{1-n}v'+p\left(x\right)v=q\left(x\right)$

Transform to  $\frac{1}{1-n}v'+p\left(x\right)v=q\left(x\right):$         $-v'+xv = e^{\left(\frac{x^2}{2}\right)}\sin \left(x\right)$

Solve $-v'+xv = e^{\left(\frac{x^2}{2}\right)}\sin \left(x\right)$ :                  $v = e^{\large\frac{x^2}{2}}cos(x) + c_1 e^{\large\frac{x^2}{2}}$

Substitute back $v = y {-1}:$       $y^{-1} = e^{\large\frac{x^2}{2}}cos(x) + c_1 e^{\large\frac{x^2}{2}}$

Isolate y :          $y = \large\frac{1}{e^{\frac{x^2}{2}}\left(\cos \left(x\right)+c_1\right)}$

The answer:     $y = \large\frac{1}{e^{\frac{x^2}{2}}\left(\cos \left(x\right)+c_1\right)}$