# Math Help - Find the solution

1. ## Find the solution

find the solution of the ivp

y' = (1-y^2)x^2
y(0) = 0

2. Originally Posted by matty888
find the solution of the ivp

y' = (1-y^2)x^2
y(0) = 0
The equation is seperatable

$\frac{dy}{dx}=(1-y^2)x^2 \iff \frac{dy}{1-y^2}=x^2dx$

All you need to do from here is integrate.

Note that

$\frac{1}{1-y^2}=\frac{\frac{1}{2}(1-y)+\frac{1}{2}(1+y)}{(1-y)(1+y)}+\frac{\frac{1}{2}}{1+y}+\frac{\frac{1}{2} }{1-y}$

3. ## More help needed

thanks empty set
i kinda have this bit
could you tell me if the answer

+- sqrt(1-e^(2/3.x^3))

sound any good????
thanks