Solve this homogeneous type equation

I can't seem to solve this question, please tell me what I am doing wrong:

**q1. **

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 + xv'

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|

x + c = -(v^2-1)

x + v^2 -1 = c

v = +-(c-x+1)

y = xv

y = +-x(c-x+1)

..

fairly sure i am correct up until i take the logs out.

The answer book says the answer is

x^2 + y^2 = cx

I am unsure how to get this.

Re: Solve this homogeneous type equation

Quote:

Originally Posted by

**99.95** I can't seem to solve this question, please tell me what I am doing wrong:

**q1. **

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 + xv'

You mean y'= v+ xv'

Quote:

1/x dx = 1/(f(v)-v) dv

First, I would have written y'= v+ xv'= (v^2- 1)/(2v)= v/2- 1/(2v) so that xv'= -1/(2v)- v/2.

Quote:

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

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

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

Okay, yes, that is what I got.

Quote:

integrating:

int(dx/x) = int(2v/-(v^2+1))

ln|x| + c = -ln|v^2-1|

You mean "ln|v^1+ 1|" don't you?

This is wrong or two reasons. First, it should be "-(v^2+ 1)" on the right side.

More importantly, e^(ln(x)+ C)= e^C e^(ln(x))= cx where c= e^C.

That is, cx= -(v^2+ 1)

Quote:

x + v^2 -1 = c

v = +-(c-x+1)

y = xv

y = +-x(c-x+1)

..

fairly sure i am correct up until i take the logs out.

The answer book says the answer is

x^2 + y^2 = cx

I am unsure how to get this.

From

Re: Solve this homogeneous type equation

Quote:

Originally Posted by

**HallsofIvy** You mean y'= v+ xv'

This is wrong or two reasons. First, it should be "-(v^2+ 1)" on the right side.

More importantly, e^(ln(x)+ C)= e^C e^(ln(x))= cx where c= e^C.

That is, cx= -(v^2+ 1)

Ah yes, how could i make such a silly mistake. That definitely makes things more easier.

Could you explain what to do from there though? I am not 100% sure.

thanks, you are a gun ivy!