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Homogenous D.E

paulmdrdo

Active member
May 13, 2013
386

please help me continue solving this,

$\displaystyle \frac{dy}{dx}=\ln(x)-\ln(y)+\frac{x-y}{x+y}$

this is where I can get to,

$\displaystyle \frac{dy}{dx}=\ln(\frac{x}{y})+\frac{x-y}{x+y}$

multiplying the 2nd term by $\frac{1}{x}$

$\displaystyle \frac{dy}{dx}=\ln(\frac{x}{y})+\frac{\frac{x}{y}-1}{\frac{x}{y}+1}$

using the substitution, $\displaystyle v=\frac{y}{x}$--->$\displaystyle y=vx$----->>$\displaystyle\frac{1}{v}=\frac{x}{y}$

$\displaystyle \frac{dy}{dx}=\ln(\frac{1}{v})+\frac{\frac{1}{v}-1}{\frac{1}{v}+1}$

until here I don't know how to proceed.. please help!
 

Fantini

"Read Euler, read Euler." - Laplace
MHB Math Helper
Feb 29, 2012
342
You have a mistake in your third equation if you really multiplied by $1/x$, because

$$\frac{x-y}{x+y} = \frac{x \left( 1 - \frac{y}{x} \right)}{x \left( 1 + \frac{y}{x} \right)} = \frac{ 1- \frac{y}{x} }{ 1 + \frac{y}{x} }.$$

Therefore with the substitution $y = vx$ we have

$$
\begin{align}
\frac{dy}{dx} & = v + \frac{dv}{dx} \\
& \text{and} \\
v + \frac{dv}{dx} &= \ln (v) + \frac{1-v}{1+v}.
\end{align}
$$

Perhaps this is more tractable. :)
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Actually, if one uses:

\(\displaystyle y=vx\)

We find by use of the product rule:

\(\displaystyle \frac{dy}{dx}=v+x\frac{dv}{dx}\)

This will lead you to a separable, but from what I can tell, an integral not expressible in terms of elementary functions.
 

paulmdrdo

Active member
May 13, 2013
386
then how can we solve this?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
then how can we solve this?
Are you certain you have copied the problem exactly as given? I could be wrong or overlooking an appropriate substitution, but I do not see how to obtain even an implicit solution, and W|A cannot either.
 

paulmdrdo

Active member
May 13, 2013
386
Are you certain you have copied the problem exactly as given? I could be wrong or overlooking an appropriate substitution, but I do not see how to obtain even an implicit solution, and W|A cannot either.
Yes, I copied it correctly.
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Yes, I copied it correctly.
The best I can do is give a solution of the form:

\(\displaystyle x=C\exp\left(\int f(v)\,dv \right)\) where $C>0$.