(b) Find a solution of the initial-value problem.

$\displaystyle

\frac {dy}{dx} = \frac {xy^3}{5} , y(0) = 4

$

I do not know how to solve a. However, This is how I solved b. Please give me a hint to solve a and check my answer for b.

(b)

$\displaystyle \frac {dy}{dx} = \frac {xy^3}{5}$

Move the ys to one side:

$\displaystyle \frac {dy}{y^3} = \frac{xdx}{5}$

Integrate:

$\displaystyle \int \frac {dy}{y^3} = \int \frac{xdx}{5}$

$\displaystyle \frac {-2}{y^2} = \frac {x^2}{10} + C$

$\displaystyle y = \sqrt{\frac{-5}{x^2}}$

I dont think my answer is correct.

Edit: Added the C.