Answer to Question #277441 in Differential Equations for Kent

Question

Find the solution of Bernoulli Equation

Given:

dy+ydx=(2xy^2)e^xdx

Answer 1

$\frac{{dy}}{{dx}} + y = 2x{y^2}{e^x} \Rightarrow y' + y = 2x{y^2}{e^x}$

Substitution:

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

Then

$u'v + uv' + uv = 2x{u^2}{v^2}{e^x}$

$u'v + u\left( {v' + v} \right) = 2x{u^2}{v^2}{e^x}$

Let

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

Then

$u'{e^{ - x}} = 2x{u^2}{e^{ - 2x}}{e^x}$

$\frac{{du}}{{dx}} = 2x{u^2} \Rightarrow \frac{{du}}{{{u^2}}} = 2xdx \Rightarrow - \frac{1}{u} = {x^2} + C \Rightarrow u = - \frac{1}{{{x^2} + C}}$

Then

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

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