We are given to solve:
Now, if we use the substitution:
We may write:
Now, back-substitute for , however, I don't see how the given result is correct.
Hello all,
I've spent the better part of an hour on a single problem. I've attempted it several ways but have yet to find the right answer - every attempt is resulting in a different answer.
The method I'm supposed to be using is substitution. The problem is (2x^2-3xy)y'=x^2+2xy-3y^2
The correct answer is y(x)=x*|ln(y^2/x)+C|.
My latest answer is y=2*x*ln(y/x) - x*ln|x| as a result of integrating (2-v)dv = -(1/x)dx where v=y/x. My previous two attempts had vastly different answers. I have a sneaking suspicion my problem is in my algebra before I start integrating.
I miss-scribed from my book - that's actually the problem right next to the problem I'm having the problem with! (2x^2-xy)y'=xy-y^2 is the problematic problem.
I believe my problem to be in the algebra - I'm rusty from not doing anything algebraically tricky for a few years.
In the problem I gave you, your second line, to the right of the second equal sign, you smack the problem upside the head with some algebra - you multiply through with 1/x^2, correct?
In my problem it appears I should do the following (Pardon my lack of Latex):
y'=(x^3+3y^3)/(2x^2-xy) * (1/x^3)/(1/x^3)
= (1+3(y^3/x^3))/((2/x)-((y/x)*(1/x))
v=y/x
y=vx
dy/dx=v+x(dv/dx)
v+x(dv/dx)=(1+3v^2)/((2/x)-(v/x) = (x^2*(3v^3+1))/(2x-v)
At about this point, I think I'm lost again. I expanded what I had and spent half a page algebraically manipulating it.
This is what I boiled it down to, likely incorrectly: 2/x-v/(x^2) + (dv/dx)*(2/v - 1) = 3v^2+1/v.