# 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 $C>0$.