# Homogeneous polar equation substitutions

• August 31st 2013, 05:10 PM
Wolvenmoon
Homogeneous polar equation substitutions
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

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.
• September 1st 2013, 05:39 AM
MarkFL
Re: Homogeneous polar equation substitutions
We are given to solve:

$\left(2x^2-3xy \right)y'=x^2+2xy-3y^2$

$\frac{dy}{dx}=\frac{x^2+2xy-3y^2}{2x^2-3xy}=\frac{1+2\left(\frac{y}{x} \right)-3\left(\frac{y}{x} \right)^2}{2-3\left(\frac{y}{x} \right)}$

Now, if we use the substitution:

$v=\frac{y}{x}\implies y=vx\implies \frac{dy}{dx}=v+x\frac{dv}{dx}$

We may write:

$v+x\frac{dv}{dx}=\frac{1+2v-3v^2}{2-3v}$

$x\frac{dv}{dx}=\frac{1+2v-3v^2}{2-3v}-v=\frac{1+2v-3v^2-v(2-3v)}{2-3v}=\frac{1}{2-3v}$

$\int 2-3v\,dv=\int\frac{1}{x}\,dx$

$2v-\frac{3}{2}v^2=\ln|x|+C$

Now, back-substitute for $v$, however, I don't see how the given result is correct.
• September 1st 2013, 05:38 PM
Wolvenmoon
Re: Homogeneous polar equation substitutions
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)

The first thing I notice about $(2x^2- xy)y'= xy- y^2$ is that all terms are of second degree- $x^2$, $y^2$, or $xy$. That suggests the substitution $u= \frac{y}{x}$ or $y= xu$. Then $y'= xu'+ u$ and the equation becomes $(2x^2- x^2u)(xu'+ u)= x^2u+ x^2u^2$. Dividing through by $x^2$ $(2- u)(xu'+ u)= (-xu+ 2x)u'+ 2u- u^2= u- u^2$ or $x(2- u)u'= -u$, which is separable: $\frac{(2- u)du}{-u}= (1- \frac{2}{u})du= \frac{dx}{x}$.