# Homogeneous 1st order Differential Equation

• Feb 19th 2011, 06:36 PM
dtwazere
Homogeneous 1st order Differential Equation
Hi guys, I've gotten so far with this but I'm stuck on what to do next:

Have I done the steps correctly? I followed what I was told but maybe I've made a mistake I cannot see.

$(2x^2+y^2)y'=2xy$

$y'= \frac{2(y/x)}{2+(y/x)^2}$

So by letting v=y/x and differentiating and substituting my value for v and dy/dx in I get the following seperable DE

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

Which I work out to be

$\frac{1}{v^2} -\ln v = \ln x +C$

But Im unsure how to isolate v, I tried using maple and It gave me an answer using the Lambert W function which I've never seen before, so I'm sure I've made a mistake along the lines..

Any help is appreciated! (Nod)
• Feb 19th 2011, 06:46 PM
mr fantastic
Quote:

Originally Posted by dtwazere
Hi guys, I've gotten so far with this but I'm stuck on what to do next:

Have I done the steps correctly? I followed what I was told but maybe I've made a mistake I cannot see.

$(2x^2+y^2)y'=2xy$

$y'= \frac{2(y/x)}{2+(y/x)^2}$

So by letting v=y/x and differentiating and substituting my value for v and dy/dx in I get the following seperable DE

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

Which I work out to be

$\frac{1}{v^2} -\ln v = \ln x +C$

But Im unsure how to isolate v, I tried using maple and It gave me an answer using the Lambert W function which I've never seen before, so I'm sure I've made a mistake along the lines..

Any help is appreciated! (Nod)

Why not just back-substitute v = y/x and use a basic log rule to simplify ....? By the way, I think the -2 should be just 2.
• Feb 19th 2011, 06:46 PM
Ackbeet
Hmm. Standard form (at least how I learned it) is

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

Substitute $x=uy,$ with $dx=u\,dy+y\,du,$ giving us

$-2uy^{2}(u\,dy+y\,du)+(2u^{2}y^{2}+y^{2})\,dy=0,$ or

$-2uy\,du+dy=0.$ Separating yields

$-2u\,du+\dfrac{dy}{y}=0.$

Can you finish? I was able to solve for $x,$ but not for $y.$
• Feb 19th 2011, 06:54 PM
Prove It
$\displaystyle (2x^2 + y^2)\,\frac{dy}{dx} = 2x\,y$

$\displaystyle \frac{2x^2 + y^2}{2x\,y} = \frac{dx}{dy}$

$\displaystyle \frac{x}{y} + \frac{y}{2x} = \frac{dx}{dy}$

$\displaystyle \frac{x}{y} + \frac{1}{2}\left(\frac{x}{y}\right)^{-1} = \frac{dx}{dy}$.

Now make the substitution $\displaystyle v = \frac{x}{y} \implies x = v\,y \implies \frac{dx}{dy} = y\,\frac{dv}{dy} + v$ and the integral becomes

$\displaystyle v + \frac{1}{2}v^{-1} = y\,\frac{dv}{dy} + v$

$\displaystyle \frac{1}{2}v^{-1} = y\,\frac{dv}{dy}$

$\displaystyle \frac{1}{2}y^{-1} = v\,\frac{dv}{dy}$

$\displaystyle \int{\frac{1}{2}y^{-1}\,dy} = \int{v\,\frac{dv}{dy}\,dy}$

$\displaystyle \frac{1}{2}\ln{|y|} + C_1 = \int{v\,dv}$

$\displaystyle \frac{1}{2}\ln{|y|} + C_1 = \frac{1}{2}v^2 + C_2$

$\displaystyle v^2 = \ln{|y|} + C$ where $\displaystyle C = 2(C_1 - C_2)$

$\displaystyle \left(\frac{x}{y}\right)^2 = \ln{|y|} + C$

$\displaystyle \frac{x^2}{y^2} = \ln{|y|} + C$

$\displaystyle x^2 = y^2(\ln{|y|} + C)$

$\displaystyle x = \pm y \sqrt{\ln{|y|} + C }$.

You will not be able to isolate $\displaystyle y$.
• Feb 19th 2011, 07:01 PM
dtwazere
implicit
Thanks for the help!!

I forgot to add it says in the question to find the implicit solution,

Would I be able to isolate it for x and then set it equal to a constant?
• Feb 21st 2011, 01:40 AM
Ackbeet
Quote:

Originally Posted by dtwazere
Thanks for the help!!

I forgot to add it says in the question to find the implicit solution,

Would I be able to isolate it for x and then set it equal to a constant?

What is "it" that you're setting equal to a constant? And why would you want to set it equal to a constant? Neither x nor y is, in this problem, in general, equal to a constant.