Question

Solve the initial-value problem y'=(y^2-3xy-5x^2)/x^2, y(1)=-1.

Answer 1

@SithsAndGiggles

Answer 2

I got y=6x/(1-Cx^6)-x
y(1)=6/(1-C)-1=-1
6/(1-C)=0
I can't find C.

Answer 3

y'=y^2/x^2-3(y/x)-5=u^2-3u-5
u+u'x=u^2-3u-5
u'x=u^2-4u-5
du/(u^2-4u-5)=dx/x
A/(u-5)+B/(u+1)=1
A=1/6, B=-1/6
(1/6)ln abs(u-5)-(1/6)ln abs(u+1)=ln abs(x)+C
ln(u-5/u+1)^(1/6)=ln abs(x)+C
((u-5)/(u+1))^(1/6)=Cx
((u-5/(u+1)=Cx^6
1-(u-5)/(u+1)=1-Cx^6
(u+1)/(u+1)-(u-5)/(u+1)=1-Cx^6
6/(u+1)=1-Cx^6
u+1=6/(1-Cx^6)

Answer 4

The answer in the book says y=-x. When I try to find C, I can't because it's undefined.

Answer 5

Hmm, I'm getting the same result... Even Wolfram can't come up with a particular solution: http://www.wolframalpha.com/input/?i=y%27%3D%28y%5E2-3xy-5x%5E2%29%2Fx%5E2%2C+y%281%29%3D-1

I think the initial value is a typo...

Answer 6

Yeah, it must be a typo then. Thanks for the help.