Solve a given initial-value problem, bernoullis equation

[shemer77]

Homework Statement

x²*(dy/dx)-2xy=3y⁴
y(1)=1/2

The Attempt at a Solution

the most I have it reduced to du/dx+2u/3x=-1/(u^8*x^2)

 

[Hiche]

This is a Bernoulli DE. Did you use the substitution $u = y^{1 - 4} = y^{-3}$? Find $dy/dx$ in terms of $du/dx$ and $y$ and substitute it in the differential equation. Eventually, you'll have a new DE of u and x and you can solve it using whatever way it should be.

 

[shemer77]

thats what i did, and when i reduced it down i got this
du/dx+2u/3x=-1/(u^8*x^2)
but I've rechecked and rechecked and i can't see if i did something wrong or what I am supposed to do next?

 

[Hiche]

What I got is different:

$dy/dx = -(1/3) y^4 (du/dx)$ ; we substitute this into the differential equation, and we get: $x^2 (-1/3) y^4 du/dx - 2xy = 3y^4$. Now divide by $x^2$ then multiply by $-3y^{-4}$ and we get: $du/dx + (6u/x) = -9/x^2$. From here, find the integrating factor and solve the DE. Your mistake is $u^8$; you multiplied by $-3y^{-4}$ so $y^4$ and $y^{-4}$ will cancel.

I might have some calculation mistakes, so I'd wait for someone else to confirm this.