Nonlinear Homogeneous Differential Equation Solution

[snowJT]

Homework Statement

I'm told that this is homogenous
(x^2-xy)y'+y^2 = 0

2. The attempt at a solution

This is going to be very painful for me to type out...

(x^2-xy)\frac{dy}{dx} = -y^2

\frac{dy}{dx} = \frac{-y^2}{(x^2-xy)}

\frac{dy}{dx} = \frac{-y^2}{(x^2-xy)}

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

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

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

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

\frac{dv}{-v^2} = \frac {dx}{x}

\int-v^-^2dv = \int\frac {dx}{x}

v^-^1 =ln|x|+C

\frac{y^-^1}{x^-^1} = ln|x|+C

y^-^1 = x ln|x|+C

\frac{1}{y} = x ln|x|+C

\frac{1}{y} = \frac {ln|x|+C}{x}

0 = \frac {y ln|x|+C}{x}

-Cx = y ln|x|

But the answer is...

xln|y| - y = Cx

 

[cristo]

Homework Statement

I'm told that this is homogenous
(x^2-xy)y'+y^2 = 0

2. The attempt at a solution

This is going to be very painful for me to type out...

(x^2-xy)\frac{dy}{dx} = -y^2

\frac{dy}{dx} = \frac{-y^2}{(x^2-xy)}

\frac{dy}{dx} = \frac{-y^2}{(x^2-xy)}

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

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

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

Try going from here again. Add another line in if you need to. Put all the v's over a common denominator before you multiply by dx.

 

[HallsofIvy]

Homework Statement

I'm told that this is homogenous
(x^2-xy)y'+y^2 = 0

2. The attempt at a solution

This is going to be very painful for me to type out...

(x^2-xy)\frac{dy}{dx} = -y^2

\frac{dy}{dx} = \frac{-y^2}{(x^2-xy)}

\frac{dy}{dx} = \frac{-y^2}{(x^2-xy)}

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

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

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

\frac{-v^2}{1-v}-v= \frac{-v^2-v+v^2}{1-v}= \frac{-v}{1-v}
The left hand side in your next line is wrong.

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

\frac{dv}{-v^2} = \frac {dx}{x}

\int-v^-^2dv = \int\frac {dx}{x}

v^-^1 =ln|x|+C

\frac{y^-^1}{x^-^1} = ln|x|+C

y^-^1 = x ln|x|+C

\frac{1}{y} = x ln|x|+C

\frac{1}{y} = \frac {ln|x|+C}{x}

0 = \frac {y ln|x|+C}{x}

-Cx = y ln|x|

But the answer is...

xln|y| - y = Cx

 

[snowJT]

Oh I understand the error, thanks, I'll work it out from there

 

[snowJT]

I've messed up again...

v+x\frac{dv}{dx} = \frac{-v^2}{(1-v)}
\frac{v(1-v)}{-v^2}dv = \frac{dx}{(x)}
\frac{v-v^2}{-v^2}dv = \frac{dx}{(x)}
\int v^-^1-1 dv = \int \frac{dx}{(x)}
ln|v|-x = ln|x|+C
ln|\frac {y}{x}|-x = ln|x|+C

and no need to go any farter.. I know its wrong...

 

[snowJT]

is it cus this line

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

should look like this?

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

 

[cristo]

Try going from this line. Simplify before you multiply by dx.

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

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

Simplify from here, and then carry on!