Solve this homogeneous type equation

nacho

Active member
Hi,
I actually made a similar thread here: Solve this homogeneous type equation
and a user pointed out a mistake in my workings, however i could still not manage to get the solution. So I was wondering if someone could help with the last few parts

Question: 2xyy' = y^2 - x^2

y' = (y^2-x^2)/(2xy)

divide through by x^2 to get:
y' = ((y/x)^2-1)/(2(y/x)) = f(y/x)

let y/x = v, thus y = v + x'v
1/x dx = 1/(f(v)-v) dv

f(v)-v= (v^2-1)/2v - v
= (-2v^2+v^2-1)/2v
= -(v^2+1)/2v

integrating:

int(dx/x) = int(2v/-(v^2+1))
ln|x| + c = -ln|v^2+1|
e^(ln(x)+ C)= e^C e^(ln(x))= cx where c= e^C
and, finally, up to the point where I am stuck:
cx= -(v^2+ 1)

The solutions say
x^2+y^2=cx

I am unsure how to get to this, any help is appreciated.

MarkFL

Staff member
We are given to solve:

$$\displaystyle 2xyy'=y^2-x^2$$

Dividing through by $$\displaystyle 2xy$$, we obtain the first order homogenous equation:

$$\displaystyle y'=\frac{1}{2}\left(\frac{y}{x}-\frac{x}{y} \right)$$

Using the substitution:

$$\displaystyle y=vx\,\therefore\,y'=v+xv'$$

we then have:

$$\displaystyle v+xv'=\frac{1}{2}\left(v-\frac{1}{v} \right)$$

$$\displaystyle xv'=-\frac{1}{2}\left(v+\frac{1}{v} \right)$$

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

Integrating, we find:

$$\displaystyle \ln\left(v^2+1 \right)=\ln\left|\frac{C}{x} \right|$$

Choosing $C$ such that $$\displaystyle 0<\frac{C}{x}$$ we may write:

$$\displaystyle v^2+1=\frac{C}{x}$$

back substitute for $v$:

$$\displaystyle \frac{y^2}{x^2}+1=\frac{C}{x}$$

multiply through by $x^2$:

$$\displaystyle x^2+y^2=Cx$$

nacho

Active member
We are given to solve:

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

Integrating, we find:

$$\displaystyle \ln\left(v^2+1 \right)=\ln\left|\frac{C}{x} \right|$$
Why is it c/x and not ln|x| + c ?

Ackbach

Indicium Physicus
Staff member
Why is it c/x and not ln|x| + c ?
$$\int\left(- \frac{1}{x} \right) dx=- \int \frac{dx}{x}=- \ln \left|x \right|+C= \ln \left|x^{-1} \right|+C= \ln \left| \frac{1}{x} \right|+C.$$
Since the $C$ is an arbitrary constant, you can simply write it as $\ln \left| C \right|$, which gives you MarkFL's result.

nacho

Active member
thanks,
just curious though. If you don't end up writing C as ln|C|, then you get a different answer in the end, don't you - Despite the fact that both are fundamentally the same thing.

This confuses me.