# Homogeneous ODE

• Dec 2nd 2011, 11:17 AM
losm1
Homogeneous ODE
Equation is $xy' = y \ln \frac{y}{x}$.
I have tried to substitute $v = \frac{y}{x}$ and I am stuck at $\frac{1}{dx}x = v (\ln v - 1)\frac{1}{dv}$

To be exact I do not know where to go from there in calculus sense.
• Dec 2nd 2011, 12:07 PM
Darkprince
Re: Homogeneous ODE
So you have (1/x)dx = -(1/(vlnv-v) )dv= -1/(v(lnv-1)

Now the integral of -1/(v(lnv-1) is -ln(lnv - 1)

Then proceed accordingly, hope I helped :)
• Dec 2nd 2011, 12:08 PM
alexmahone
Re: Homogeneous ODE
Quote:

Originally Posted by losm1
Equation is $xy' = y \ln \frac{y}{x}$.
I have tried to substitute $v = \frac{y}{x}$ and I am stuck at $\frac{1}{dx}x = v (\ln v - 1)\frac{1}{dv}$

To be exact I do not know where to go from there in calculus sense.

$\frac{dv}{v(\ln v-1)}=\ln (\ln v-1)+C$
• Dec 3rd 2011, 03:19 AM
losm1
Re: Homogeneous ODE
In my example differentials dx and dv are in denominator:

$\frac{1}{dx}x = v (\ln v - 1)\frac{1}{dv}$

I'm having trouble applying your formula in this case. Can you please clarify further?
• Dec 3rd 2011, 04:13 AM
Prove It
Re: Homogeneous ODE
Quote:

Originally Posted by losm1
Equation is $xy' = y \ln \frac{y}{x}$.
I have tried to substitute $v = \frac{y}{x}$ and I am stuck at $\frac{1}{dx}x = v (\ln v - 1)\frac{1}{dv}$

To be exact I do not know where to go from there in calculus sense.

Make the substitution \displaystyle \begin{align*} v = \frac{y}{x} \implies y = v\,x \implies \frac{dy}{dx} = \frac{dy}{dx} = v + x\,\frac{dv}{dx} \end{align*} and the DE becomes

\displaystyle \begin{align*} x\,\frac{dy}{dx} &= y\ln{\left(\frac{y}{x}\right)} \\ x\left(v + x\,\frac{dv}{dx}\right) &= v\,x\ln{v} \\ v + x\,\frac{dv}{dx} &= v\ln{v} \\ x\,\frac{dv}{dx} &= v\ln{v} - v \\ x\,\frac{dv}{dx} &= v\left(\ln{v} - 1\right) \\ \frac{1}{v\left(\ln{v} - 1\right)}\,\frac{dv}{dx} &= \frac{1}{x} \\ \int{\frac{1}{v\left(\ln{v} - 1\right)}\,\frac{dv}{dx}\,dx} &= \int{\frac{1}{x}\,dx} \\ \int{\frac{1}{\ln{v} - 1}\,\frac{1}{v}\,dv} &= \ln{|x|} + C_1 \\ \int{\frac{1}{u}\,du} &= \ln{|x|} + C_1 \textrm{ after making the substitution }u = \ln{v} - 1 \implies du = \frac{1}{v}\,dv \\ \ln{|u|} + C_2 &= \ln{|x|} + C_1 \\ \ln{|u|} - \ln{|x|} &= C \textrm{ where }C = C_1 - C_2 \\ \ln{\left|\frac{u}{x}\right|} &= C \\ \ln{\left|\frac{\ln{v} - 1}{x}\right|} &= C \\ \ln{\left|\frac{\ln{\left(\frac{y}{x}\right)} - 1}{x}\right|} &= C \\ \frac{\ln{\left(\frac{y}{x}\right)} - 1}{x} &= A \textrm{ where } A = \pm e^C\end{align*}

You could get y in terms of x if you wanted to.
• Dec 3rd 2011, 05:17 AM
Darkprince
Re: Homogeneous ODE
Quote:

Originally Posted by losm1
In my example differentials dx and dv are in denominator:

$\frac{1}{dx}x = v (\ln v - 1)\frac{1}{dv}$

I'm having trouble applying your formula in this case. Can you please clarify further?

So you have x/dx = v(lnv - 1)/dv implies xdv = v(lnv-1)dx implies (1/x)dx = (1/v(lnv-1)) dv
• Dec 3rd 2011, 09:47 AM
tom@ballooncalculus
Re: Homogeneous ODE
Yes, that is the idea. However, just in case an overview helps...

http://www.ballooncalculus.org/draw/double/five.png
http://www.ballooncalculus.org/draw/double/fivea.png
http://www.ballooncalculus.org/draw/double/fiveb.png

... where (key in spoiler) ...

Spoiler:
http://www.ballooncalculus.org/asy/chain.png

... is the chain rule. Straight continuous lines differentiate downwards (integrate up) with respect to the main variable (in this case x), and the straight dashed line similarly but with respect to the dashed balloon expression (the inner function of the composite which is subject to the chain rule).

__________________________________________________ __________

Don't integrate - balloontegrate!

Balloon Calculus; standard integrals, derivatives and methods

Balloon Calculus Drawing with LaTeX and Asymptote!