Question

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

Answer 1

z=1-ye^(-2x)
dz/dx=-y*-2e^(-2x)+e^(-2x)(-dy/dx)
dz/dx=2e^(-2x)*y-e^(-2x)*dy/dx
dz/dx=e^(-2x)*(2y-1)*dy/dx
dz/dx=((2y-1)/(e^(2x)))*dy/dx
dy/dx=dz/dx*(e^2x)/(2y-1)
ye^(-2x)=1-z
y=e^(2x)(1-z)

Answer 2

When I plug y into the whole equation, it's just too much to simplify.

Answer 3

@Idealist10 isn't this the equation we worked on a few days ago?

Answer 4

I think that's the other equation.

Answer 5

This is the question I'm referring to: http://openstudy.com/users/idealist10#/updates/53f63ae7e4b0f0e909e08dbf

But yeah, you're right, this DE is slightly different. We can probably used that partial fractions method I stumbled upon to find a fundamental solution. We have the same problem of only getting one fundamental solution from the homogeneous part of the equation.

Answer 6

I'm also not sure I follow the reasoning for the work you've done so far. Care to explain?

Answer 7

So what I did was that I did z substitution, see my work, the first step is z=1-ye^(-2x) and I found the derivative of z, dz/dx,

Answer 8

Okay, so that substitution would give
\[z=1-ye^{-2x}~~\iff~~y=\frac{1-z}{e^{-2x}}\]
Differentiating gives
\[\begin{align*}z'&=2e^{-2x}y-e^{-2x}y'\\
y'&=\frac{2e^{-2x}y-z'}{e^{-2x}}\\
y'&=\frac{2e^{-2x}y}{e^{-2x}}-\frac{z'}{e^{-2x}}\\
y'&=2y-\frac{z'}{e^{-2x}}\\
y'-2y&=-e^{2x}z'
\end{align*}\]
So the equation is now
\[-e^{2x}z'=\frac{xe^{2x}}{z}\]
Interesting... I never would have thought we could make this a separable equation! Simplifying a bit, you have
\[z~dz=-x~dx\]

Answer 9

Yeah, this problem is interesting. I guess it's because the direction says 'followed by separation of variables'.

Answer 10

I think those directions mean they want you to solve the equation in two different ways. I don't see how you can solve a DE by two methods at the same time (substitution doesn't count because that only gives you another DE that's more accommodating).

Answer 11

I see, you made this problem simpler! You're awesome!