Question

Solve the following differential equation:

y' - 2xy = 2xe^(x^2)

Answer 1

Differential equation:
\[y' -2xy = 2xe ^{x ^{2}}\]


can some one check if this answer is correct?

y = x^2*e^\[y = x ^{2}e ^{x ^{2}} + ce ^{x ^{2}}\]

Answer 2

The integrating factor is u=e^(-x^2) so

(uy)' = 2x
uy = x^2 + C
y = x^2 . e^(x^2) + c.e^(x^2)

So, yep. Looks right to me.

Answer 3

*yeaahhh i finally got a good answer =D

Answer 4

Thank you somuch JamesJ :)

Answer 5

np.