# Math Help - Where did I make a mistake? (1st order homogeneous ODE)

1. ## Where did I make a mistake? (1st order homogeneous ODE)

I have a homogeneous ODE which I'm trying to solve:

$\frac{dy}{dx} = \frac{x^2 + y^2}{xy - x^2}$

$c(x + y)^2 = xe^{\frac{y}{x}}$

Here's what I'm doing:

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

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

$\int \frac{1}{x} dx = \int \frac{v - 1}{v + 1} dv$

$ln(x) = \int 1 - \frac{2}{v + 1} dv$

$ln(x) = v - 2ln(v + 1) + c$

$x = \frac{e^v}{(v + 1)^2} c$

$x = \frac{e^{\frac{y}{x}}}{(\frac{y}{x} + 1)^2} c$

$cx = \frac{e^{\frac{y}{x}}}{(\frac{y + x}{x})^2}$

...and this obviously looks nothing like the solution.

Where did I make a mistake?

2. ## Re: Where did I make a mistake? (1st order homogeneous ODE)

Originally Posted by Lancet
I have a homogeneous ODE which I'm trying to solve:

$\frac{dy}{dx} = \frac{x^2 + y^2}{xy - x^2}$

$c(x + y)^2 = xe^{\frac{y}{x}}$

Here's what I'm doing:

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

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

$\int \frac{1}{x} dx = \int \frac{v - 1}{v + 1} dv$

$ln(x) = \int 1 - \frac{2}{v + 1} dv$

$ln(x) = v - 2ln(v + 1) + c$

$x = \frac{e^v}{(v + 1)^2} c$

$x = \frac{e^{\frac{y}{x}}}{(\frac{y}{x} + 1)^2} c$

$cx = \frac{e^{\frac{y}{x}}}{(\frac{y + x}{x})^2}$

...and this obviously looks nothing like the solution.

Where did I make a mistake?

I'ts the same!

$cx = \frac{e^{\frac{y}{x}}}{(\frac{y + x}{x})^2}$

$cx(\frac{y + x}{x})^2 = {e^{\frac{y}{x}}}$

$cx(\frac{(y + x)^2}{x^2}) = {e^{\frac{y}{x}}}$

$c(\frac{(y + x)^2}{x}) = {e^{\frac{y}{x}}}$

$(\frac{(y + x)^2}{x}) = \frac{{e^{\frac{y}{x}}}}{c}$

$(y + x)^2 = x\frac{{e^{\frac{y}{x}}}}{c}$

\frac{1}{c}=C

$(y + x)^2 = Cx{e^{\frac{y}{x}}}$

3. ## Re: Where did I make a mistake? (1st order homogeneous ODE)

Originally Posted by Also sprach Zarathustra
I'ts the same!
That was a pleasant surprise!

How do you get there from here, if you don't mind my asking?

4. ## Re: Where did I make a mistake? (1st order homogeneous ODE)

Originally Posted by Also sprach Zarathustra
I'ts the same!

$cx = \frac{e^{\frac{y}{x}}}{(\frac{y + x}{x})^2}$

$cx(\frac{y + x}{x})^2 = {e^{\frac{y}{x}}}$

$cx(\frac{(y + x)^2}{x^2}) = {e^{\frac{y}{x}}}$

$c(\frac{(y + x)^2}{x}) = {e^{\frac{y}{x}}}$

$(\frac{(y + x)^2}{x}) = \frac{{e^{\frac{y}{x}}}}{c}$

$(y + x)^2 = x\frac{{e^{\frac{y}{x}}}}{c}$

\frac{1}{c}=C

$(y + x)^2 = Cx{e^{\frac{y}{x}}}$
In the 3rd last line of his post he has $(y + x)^2 = x\frac{{e^{\frac{y}{x}}}}{c}$

which is the same as $c(y + x)^2 = xe^{\frac{y}{x}}$ ie your books answer.

5. ## Re: Where did I make a mistake? (1st order homogeneous ODE)

I understand now! Thanks, guys!