Question

Solve...
y' - 2xy = e^(x^2)

Answer 1

i ended up with ... (x+c)/(e^(-x^2)) = y

Answer 2

Well, that might be a solution if x 0 or x = -c, but with only two points of any interest, this seems unsatisfactory.

How did you manage that response?

Answer 3

btw - Why would you EVER write \(\dfrac{x+c}{e^{-x^{2}}}\), when you could write \((x+c)e^{x^{2}}\)?

Answer 4

ah that is true... thats the answer i was given... but didnt know how to get it

Answer 5

I might just be tempted to guess on thhis one.

Try y = Ae^(x^2). If that result proves unsatisffaactorry,

Try y = Ae^(x^2)+ Bxe^{x^2}. If yoou still don't like it,, keep adding an x and a parameter until you do like it.

Answer 6

I think finding an integrating factor might require less work:
\[y'-2xy=e^{x^2}\]
gives you an integrating factor of
\[\mu(x)=e^{\int(-2x)~dx}=e^{-x^2}\]
So
\[e^{-x^2}y'-2xe^{-x^2}y=e^{-x^2}e^{x^2}\\
\frac{d}{dx}\left[e^{-x^2}y\right]=1\\
e^{-x^2}y=\int dx\\
y=\frac{x+C}{e^{-x^2}}=(x+C)e^{-x^2}\]

Answer 7

Oh, okay. Fine. It's still +x^2 in the numerator.

Answer 8

\(\color{blue}{\text{Originally Posted by}}\) @SithsAndGiggles
\[y=\frac{x+C}{e^{-x^2}}=(x+C)e^{x^2}\]
\(\color{blue}{\text{End of Quote}}\)
Yep, small typo :)