differential equation integrating to a natural log?

Question

i'll write what i've done so far but i'm stuck at the end at the *****

so the problem was dy/dx= (x^2 + y x^2) / 3

and i separated the variables and got
(3 dy) / (1+y) = x^2 dx

i integrated both sides and got
3 ln(1+y) + C = (x^3)/3 + C

with the C brought over to the x side, it's still
3 ln(1+y) = (x^3)/3 + C

i need y alone, with no ln, so i raised what was written to be exponants of an e to cancel the ln's
e^ (3 ln(1+y)) = e^ ((x^3)/3 + C)

***********

i can simplify the x side to C e^((x^3)/3), 
but what about the y side??

i have no idea because the 3 is there.  without the 3, it'd be just 1+y.
so what do i do now?

δοτζο · Accepted Answer

3 ln(1+y) = ln((1+y)³)  

How about now?

jonny · Answer

Go back to when you were separating your variables.  You could leave the 3 on the right side with the x's.  So you'd get dy/(1+y) = x^2/3 *dx

Now integrating both sides you'll get:  ln(1+y) = x^3 + C

Now raise both with the e:  e^(ln(1+y)) = e^(x^3 + C) = e^(x^3)*e^C :  say e^C = A

This becomes:  1 + y = Ae^x^3

Subtract 1 on both sides to get your final answer:  y = Ae^(x^3) - 1

Michael · Answer

3 ln(1+y) = (x^3)/3 + C
divide both sides by 3 before raising them as exponents of e.
So ln(1+y) = (x^3)/9 + C
Now you will get 1 + y = C e^((x^3)/9), which means that you would have y = C e^((x^3)/9) - 1, but it looks better to me if it is y = -1 + Ce^((x^3)/9) because the C stuff is at the end

rupesh · Answer

it wud b e^(ln(1+y)^3) as rule of logarithm so it will finally reduce to (1+y)^3

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