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 [itex]u = y^{1 - 4} = y^{-3}[/itex]? Find [itex]dy/dx[/itex] in terms of [itex]du/dx[/itex] and [itex]y[/itex] 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:

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

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