Homogenous D.E

paulmdrdo

Active member

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

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

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

Fantini

MHB Math Helper
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

Staff member
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
then how can we solve this?

MarkFL

Staff member
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
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

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