Question

Solve y'=(y^2)(e^-x)+4y+2e^x explicitly.

Answer 1

@amistre64 @ganeshie8 @iambatman @myininaya

Answer 2

lookup bernouli equations

Answer 3

I know the bernouli method but I don't know what v should be.

Answer 4

wait, this is not exactly bernouli

y'-4y = (y^2)(e^-x)+2e^x

Answer 5

we can't factor out y^2 on right hand side so this is not bernouli yet

Answer 6

So what should we do?

Answer 7

@myininaya @iambatman @ash2326

Answer 8

Hint: it's a quadratic in disguise.

Answer 9

The right side, I mean.

Answer 10

Well maybe it's not exactly a hint, just an observation :P

Answer 11

There are some initial steps you can take to make this a semi-homogeneous equation. Let \(u=e^x\). From the original eq, factor out \(e^x\):
\[\begin{align*}
y'&=e^x\left(y^2e^{-2x}+4ye^{-x}+2\right)\\\\
y'&=u\left(\left(\frac{y}{u}\right)^2+4\frac{y}{u}+2\right)
\end{align*}\]
Let \(y=ut\), so \(y'=ut'+t\):
\[\begin{align*}
ut'+t&=u\left(t^2+4t+2\right)\\\\
t'&=t^2+\left(4-\frac{1}{u}\right)t+2
\end{align*}\]
Hmm...

Answer 12

Then what would you do?

Answer 13

http://math.stackexchange.com/questions/947593/solve-y-y2e-x4y2ex

Answer 14

I should have stuck with my first intuition... @Idealist10 you have everything you need to solve with ganeshie's link.

Answer 15

Thank you!