1. ## Differential Equation 1

Find the solution:

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

So far I have:

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

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

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

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

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

2. Originally Posted by ahhh
Find the solution:

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

So far I have:

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

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

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

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

No matter what I do, I'm just going in circles. I can't seem to get the variables separated.
$\displaystyle \frac{\partial }{\partial y}(xy)=x$ and

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

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

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

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

This is now an exact equation so we get

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

Now taking the partial with respect to y we get

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

but this must equal

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

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

so $\displaystyle \frac{1}{2}x^2y^{-2}-\ln(y)=c$ is an implicit solution

3. 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 $\displaystyle y=vx$ and $\displaystyle dy=vdx+xdv$.

4. Originally Posted by ahhh
Find the solution:

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

So far I have:

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

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

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

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

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...

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

simplifying

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

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

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

I hope this helps...

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

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

simplifying

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

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

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

I hope this helps...
Thank you so much! I couldn't seem to get the variables separated.

6. So it should be $\displaystyle \frac {[\frac {y}{x}]^{-2}}{-2} + \ln \frac {y}{x} = \ln \frac {1}{x}+c$

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

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

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

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

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

Yay!!

8. Oh wait...dumb question, but how did you simplify
$\displaystyle \frac {[\frac {y}{x}]^{-2}}{-2}$

into this

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

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

into this

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

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