MHB Bernoulli Equation Checking the solution

[cbarker1]

Dear Everyone,

I need someone to check the solution,

The question identity and then solve,

The Equation is Bernoulli
To Solve:

$(ye^{-2x}+y^3)dx-e^{-2x}dy=0$

$\frac{1}{dx}(ye^{x}+y^3)dx=\frac{1}{dx}(e^{2x}dy)$

$ye^{-2x}+y^3=e^{-2x}\d{y}{x}$

$e^{-2x}y^{\prime}-ye^{-2x}=y^3$

Let $v={y}^{1-3}$

$v={y}^{-2}\implies {v}^{-\frac{1}{2}}=y$

$-\frac{1}{2}{v}^{-\frac{3}{2}}v^{\prime}=y^{\prime}$

$-\frac{1}{2}{v}^{-\frac{3}{2}}e^{-2x}v^{\prime}-e^{-2x}{v}^{-\frac{1}{2}}={\left\{{v}^{-\frac{1}{2}}\right\}}^{3}\implies-\frac{1}{2}{v}^{-\frac{3}{2}}e^{-2x}v^{\prime}-e^{-2x}{v}^{-\frac{1}{2}}={v}^{-\frac{3}{2}}$

${v}^{\frac{3}{2}}\left\{-\frac{1}{2}{v}^{-\frac{3}{2}}e^{-2x}v^{\prime}-e^{-2x}{v}^{-\frac{1}{2}}\right\}={v}^{-\frac{3}{2}}{v}^{\frac{3}{2}}$

$e^{-2x}v^{\prime}+2e^{-2x}v=-2$

$e^{2x}\left\{e^{-2x}v^{\prime}+2e^{-2x}v\right\}=-2e^{2x}$

$v^{\prime}+2v=-2e^{2x}$
Finding the Integrating Factor:

$\mu(x)=e^{\int2dx}=e^{2x}$

$e^{2x}v^{\prime}+2e^{2x}v=-2e^{2x}e^{2x}$

$\left[e^{2x}v\right]^{\prime}=-2e^{4x}$

$\int\left[e^{2x}v\right]^{\prime}=\int-2e^{4x}dx$

$e^{2x}v=-\frac{1}{2}e^{4x}+C$

$v=-\frac{1}{2}e^{2x}+Ce^{-2x}$

${y}^{-2}=-\frac{1}{2}e^{2x}+Ce^{-2x}$

 

[Ackbach]

The cool thing about solving DE's is that you always have a good, easy, reliable way to check your answer: plug back into the DE. In your case, the Mathematica code

y[x_]=(-(1/2)*Exp[2x]+C[1] Exp[-2x])^(-1/2) 
Simplify[y[x] Exp[-2x]+(y[x])^3-Exp[-2x]y'[x]]

produced a zero, so I would say you certainly found a solution!