Answer to Question #141660 in Differential Equations for Nikhil

Question

Solve
dy/dx + xy=y^2×e^(x^2/2)×cosx

Answer 1

$y' + xy = {y^2}{e^{\frac{{{x^2}}}{2}}}\cos x$

Let

$y = uv \Rightarrow y' = u'v + uv'$

Then

$\begin{array}{l} u'v + uv' + xuv = {u^2}{v^2}{e^{\frac{{{x^2}}}{2}}}\cos x\\ u'v + u\left( {v' + xv} \right) = {u^2}{v^2}{e^{\frac{{{x^2}}}{2}}}\cos x \end{array}$

Let

$v' + xv = 0 \Rightarrow \frac{{dv}}{{dx}} = - xv \Rightarrow \frac{{dv}}{v} = - xdx \Rightarrow \ln v = - \frac{{{x^2}}}{2} \Rightarrow v = {e^{ - \frac{{{x^2}}}{2}}}$

Then

$u'{e^{ - \frac{{{x^2}}}{2}}} = {u^2}{e^{ - {x^2}}}{e^{\frac{{{x^2}}}{2}}}\cos x$

$\begin{array}{l} u' = {u^2}{e^{ - {x^2}}}{e^{{x^2}}}\cos x\\ u' = {u^2}\cos x \end{array}$

$\begin{array}{l} \frac{{du}}{{dx}} = {u^2}\cos x\\ \frac{{du}}{{{u^2}}} = \cos xdx\\ - \frac{1}{u} = \sin x - C\\ u = \frac{1}{{C - \sin x}} \end{array}$

$y = uv = \frac{{{e^{ - \frac{{{x^2}}}{2}}}}}{{C - \sin x}}$

Answer: $y = \frac{{{e^{ - \frac{{{x^2}}}{2}}}}}{{C - \sin x}}$