Differential Equation of Order 1

[verticos]

Homework Statement

Find the particular solution to the initial value problem: (3xy - 4x - 1)dy + y(y - 2)dx = 0; when x=1, y=2

Homework Equations

dy + (p(x)y - q(x))dx = 0
e^(-∫p(x)dx) * (c + ∫e^(∫p(x)dx) * q(x)dx)

The Attempt at a Solution

Sorry if this is vague, but I just spent 30 min typing out the entire process only to have it deleted. This is where I got stuck:

p(y) = 3y - 4 / y(y - 2)
q(y) = 1 / y(y - 2)

x = [y^(1/2) / (y-2)^(1/2)] * (c + ∫dy/[y^(3/2) * (y - 2)^(1/2)])

I don't know how to do the remaining integral. I think it's partial fractions but the y^(3/2) is confusing me.

 

[ehild]

I can not follow you. If you found the integrating factor which is simply y, (Read: http://www.math.hmc.edu/calculus/tutorials/odes/) you get the exact equation

y^2(y-2)dx+y(3xy-4x-1)dy.

You find the potential function U(x,y) by integrating y^2(y-2) with respect to x and adding an "integration constant" in the form g(y), or integrating y(3xy-4x-1) with respect to y, and including the integration constant f(x). Find f(x) and g(y) so that both forms of U(x,y) are identical. The solution is U(x,y)=constant. Substitute the initial condition to get the appropriate value of the constant.

ehild