Homogenous D.E

[paulmdrdo]

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]

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]

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]

then how can we solve this?

 

[MarkFL]

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]

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]

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 $C>0$.