Question

If dy/dx=1/(y+x^2) and y(0)=2, find an equation for y in terms of x.

Answer 1

Sorry, I mean
dy/dx = 1/(y+x^2y)

Answer 2

I know that I have to separate the variables.
So (y+x^2y)dy=dx. I don't know what to do next.

Answer 3

factor out y

Answer 4

is this right?
ydy=dx/1+x^2

Answer 5

yeah

Answer 6

ok now is this next step correct, after i do the integral?
y^2/2 = ln|1+x^2|

Answer 7

try arctan(x)

Answer 8

what???

Answer 9

the integral of 1/(1+x^2) is arctan(x)

Answer 10

oh. why can't i use the rule int(du/u)=ln|u|+c?

Answer 11

du = 2x dx, but you don't have an extra x on the numerator

Answer 12

ok so then i find c right? which is 2. and i plug it in and get \[y=\sqrt{2(arctanx+2)}\]