1. ## Homogenous D.E

Any help solving homogenous differntial equation would be greatly appreciated:

dy/dx=(-2x^2+2y^2)/xy

The following is what i've worked out (could be incorrect):
dy/dx=2(y/x-x/y)
let v=y/x, dy/dx=v+x*dv/dx
v+x*dv/dx=2(v-1/v), x*dv/dx=2(v-1/v)-v
therefore, x*dv/dx=(v^2-2)/v

1/x dx=v/(v^2-2) dv
-ln(x)+C=1/2 ln(v^2-2)
C/x=1/2(y^2/x^2-2)
C/x=y^2/2x^2-1
C/x+1=y^2/2x^2
y^2=2x^2(C/x+1)
y=sqrt(2x^2(C/x+1))

I must of gone wrong somewhere as it says the answer is incorrect, i dont know where though...

2. Originally Posted by skirk34
Any help solving homogenous differntial equation would be greatly appreciated:

dy/dx=(-2x^2+2y^2)/xy

The following is what i've worked out (could be incorrect):
dy/dx=2(y/x-x/y)
let v=y/x, dy/dx=v+x*dv/dx
v+x*dv/dx=2(v-1/v), x*dv/dx=2(v-1/v)-v
therefore, x*dv/dx=(v^2-2)/v

1/x dx=v/(v^2-2) dv
-ln(x)+C=1/2 ln(v^2-2)
C/x=1/2(y^2/x^2-2)
C/x=y^2/2x^2-1
C/x+1=y^2/2x^2
y^2=2x^2(C/x+1)
y=sqrt(2x^2(C/x+1))

I must of gone wrong somewhere as it says the answer is incorrect, i dont know where though...
$\int 1/x dx=\int v/(v^2-2) dv$

$\ln(Ax)= \frac{1}{2} \ln(v^2-2)$

$A$ is an arbitary constant, and whatever transformation is applied to it I will use the same symbol for the result, so it is not the same thing in every expression that follows:

$\ln(Ax^2)= \ln(v^2-2)$

so:

$Ax^2=v^2-2$

$v^2=Ax^2+2$

$y^2=x^2(Ax^2+2)$

etc

RonL

3. cool, so it was just a simple mistake i made in evaluating the logarithms. Thxs

4. Originally Posted by skirk34
cool, so it was just a simple mistake i made in evaluating the logarithms. Thxs
No look at the sign of the integral of 1/x (Note there is also an implicit assumption that x is positive as otherwise the log should be of A|x|)

RonL