Solve y'+y=(2xe^-x)/(1+ye^x) using variation of parameters followed by separation of variables.

Question

idealist10 · Answer

@SithsAndGiggles

idealist10 · Answer

dy/dx=(2xe^-x)/(1+ye^x)-y?

anonymous · Answer

So you have to solve the same equation twice, using a different method each time?

idealist10 · Answer

Same equation twice? I don't think so...

anonymous · Answer

I guess I just don't understand what the directions mean by "...followed by..."

Anyway, to use VoP, you'll need two known fundamental solutions. From the homogeneous equation
$$y'+y=0$$
you have the characteristic equation
$$r+1=0~~\iff~~r=-1$$
So one solution is $C_1e^{-x}$ (the fundamental solution is then $e^{-x}$). Not sure about the second solution just yet...

idealist10 · Answer

I'll come back. Let me look for problems like this.

anonymous · Answer

I don't know if it's relevant or helpful, but I stumbled across an interesting way to express the right hand side in terms of partial fractions:
$$\begin{align*}\frac{xe^{-x}}{1+ye^x}&=\frac{x}{e^x(1+ye^x)}\\
&=\frac{f(x,y)}{e^x}+\frac{g(x,y)}{1+ye^x}\\
x&=f(x,y)(1+ye^x)+g(x,y)e^x\\
x&=f(x,y)+e^x\bigg(yf(x,y)+g(x,y)\bigg)
\end{align*}$$
which means $f(x,y)=x$ and $yf(x,y)+g(x,y)=0$. Substituting gives $g(x,y)=-xy$, and so
$$RHS=\frac{xe^{-x}}{1+ye^x}=\frac{x}{e^x}-\frac{xy}{1+ye^x}$$

anonymous · Answer

So now you can split up the equation into two non-homogeneous ones:
$$y'+y=xe^{-x}~~~~~~~~~~~\text{(a simple linear equation)}\
y'+y=-\frac{xy}{1+ye^x}~~\text{(not quite sure yet)}$$

anonymous · Answer

Mathematica 9 came up with the follwoing:$$y(x)=-e^{-x}\pm e^{-2 x} \sqrt{c_1 e^{2 x}+2 e^{2 x} x^2+e^{2 x}}  $$

anonymous · Answer

The linear equation gives $e^{-x}x^2$ as a fundamental solution (I'll leave the details of finding it to you). This solution shows up under the square root in rob's comment

idealist10 · Answer

Thanks for the extra info.

anonymous · Answer

You're welcome!

So when you use VoP, you'll be using $y_1=e^{-x}$ and $y_2=x^2e^{-x}$, which should be linearly independent... Check to make sure that's true.

anonymous · Answer

I'm not sure where the separation of variables comes in though...