# Differential Equation 1

• March 11th 2009, 07:15 PM
ahhh
Differential Equation 1
Find the solution:

$xy dx - (x^{2}+y^{2}) dy =0$

So far I have:

$y=vx$
$dy=vdx+xdv$

Plug the $y$ and $dy$ into the equation to get:

$x(vx) dx -(x^{2}+(vx)^{2}) (vdx+xdv)=0$

$vx^{2} dx -[vx^{2} dx+ x^{3} dv+ v^{3}x^{2}dx+v^{2}x^{3}dv]=0$

No matter what I do, I'm just going in circles. I can't seem to get the variables separated.
• March 11th 2009, 07:35 PM
TheEmptySet
Quote:

Originally Posted by ahhh
Find the solution:

$xy dx - (x^{2}+y^{2}) dy =0$

So far I have:

$y=vx$
$dy=vdx+xdv$

Plug the $y$ and $dy$ into the equation to get:

$x(vx) dx -(x^{2}+(vx)^{2}) (vdx+xdv)=0$

$vx^{2} dx -[vx^{2} dx+ x^{3} dv+ v^{3}x^{2}dx+v^{2}x^{3}dv]=0$

No matter what I do, I'm just going in circles. I can't seem to get the variables separated.

$\frac{\partial }{\partial y}(xy)=x$ and

$\frac{\partial }{\partial x}(-(x^2+y^2))=-2x$

$\frac{-2x-x}{xy}=-\frac{3}{y}$

So the integrating factor is $e^{\int -\frac{3}{y}dy}=y^{-3}$

$xy^{-2}dx-(x^2y^{-3}+y^{-1})dy=0$

This is now an exact equation so we get

$\frac{\partial f}{\partial x}=xy^{-2} \implies f(x,y)=\frac{1}{2}x^2y^{-2}+g(y)$

Now taking the partial with respect to y we get

$\frac{\partial f}{\partial y}=-x^2y^{-3}+g'(y)$

but this must equal

$-(x^2y^{-3}+y^{-1})=-x^2y^{-3}+g'(y)$

$-y^{-1}=g'(y) \implies g(y)=-\ln(y)$

so $\frac{1}{2}x^2y^{-2}-\ln(y)=c$ is an implicit solution
• March 11th 2009, 07:40 PM
ahhh
I'm sorry, but I don't understand the process. In my class, the only way we have solved differentials so far is to use $y=vx$ and $dy=vdx+xdv$.
• March 11th 2009, 07:48 PM
TheEmptySet
Quote:

Originally Posted by ahhh
Find the solution:

$xy dx - (x^{2}+y^{2}) dy =0$

So far I have:

$y=vx$
$dy=vdx+xdv$

Plug the $y$ and $dy$ into the equation to get:

$x(vx) dx -(x^{2}+(vx)^{2}) (vdx+xdv)=0$

$vx^{2} dx -[vx^{2} dx+ x^{3} dv+ v^{3}x^{2}dx+v^{2}x^{3}dv]=0$

No matter what I do, I'm just going in circles. I can't seem to get the variables separated.

ummm sorry lets try this again...

$vx^{2} dx -[vx^{2} dx+ x^{3} dv+ v^{3}x^{2}dx+v^{2}x^{3}dv]=0$

simplifying

$-x^3dv-v^{3}x^2dx-v^{2}x^3dv=0$

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

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

I hope this helps...
• March 11th 2009, 08:06 PM
ahhh
Quote:

Originally Posted by TheEmptySet
ummm sorry lets try this again...

$vx^{2} dx -[vx^{2} dx+ x^{3} dv+ v^{3}x^{2}dx+v^{2}x^{3}dv]=0$

simplifying

$-x^3dv-v^{3}x^2dx-v^{2}x^3dv=0$

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

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

I hope this helps...

Thank you so much! I couldn't seem to get the variables separated.
• March 11th 2009, 08:14 PM
ahhh
So it should be $\frac {[\frac {y}{x}]^{-2}}{-2} + \ln \frac {y}{x} = \ln \frac {1}{x}+c$

I'm not worried about simplifying it because I am liable to mess up doing algebra.
• March 11th 2009, 08:24 PM
TheEmptySet
Quote:

Originally Posted by ahhh
So it should be $\frac {[\frac {y}{x}]^{-2}}{-2} + \ln \frac {y}{x} = \ln \frac {1}{x}+c$

I'm not worried about simplifying it because I am liable to mess up doing algebra.

Just for fun and to note if we do simplify a little we can show that this is the same as the solution we got solving the exact equation

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

$\frac{1}{2}x^2y^{-2}-\ln(y)=c$

Yay!!
• March 11th 2009, 08:33 PM
ahhh
Oh wait...dumb question, but how did you simplify
$\frac {[\frac {y}{x}]^{-2}}{-2}
$

into this

$\frac{1}{2}x^2y^{-2}$
• March 11th 2009, 09:13 PM
TheEmptySet
Quote:

Originally Posted by ahhh
Oh wait...dumb question, but how did you simplify
$\frac {[\frac {y}{x}]^{-2}}{-2}$ $
$

into this

$\frac{1}{2}x^2y^{-2}$

$\frac {[\frac {y}{x}]^{-2}}{-2}=\frac{\left( \frac{x}{y} \right)^2}{\frac{-2}{1}}=\left( \frac{x}{y} \right)^2 \left( \frac{1}{-2} \right) =\frac{x^2}{-2y^2}=-\frac{1}{2}x^2y^{-2}$

Then I multiplied the whole equation by -1 to change it from a negative to a +.