How do I solve the differential equation: $ \frac{dy}{dx} = \frac{x+y}{x-y} $? $$ \frac{dy}{dx} = \frac{x+y}{x-y}   $$
I have tried this problem so long... such as 
please help..
 A: Well, we have that:
$$\text{y}'\left(x\right)=\frac{x+\text{y}\left(x\right)}{x-\text{y}\left(x\right)}\tag1$$
Let $x\cdot\text{r}\left(x\right)=\text{y}\left(x\right)$:
$$x\cdot\text{r}'\left(x\right)+\text{r}\left(x\right)=\frac{x+x\cdot\text{r}\left(x\right)}{x-x\cdot\text{r}\left(x\right)}\space\Longleftrightarrow\space\int-\frac{\text{r}'\left(x\right)\cdot\left(\text{r}\left(x\right)-1\right)}{1+\text{r}\left(x\right)^2}\space\text{d}x=\int\frac{1}{x}\space\text{d}x\tag2$$
Now, use:


*

*For the LHS, substitute $\text{u}=\text{r}\left(x\right)$:
$$\int-\frac{\text{r}'\left(x\right)\cdot\left(\text{r}\left(x\right)-1\right)}{1+\text{r}\left(x\right)^2}\space\text{d}x=\int\frac{1-\text{u}}{1+\text{u}^2}\space\text{d}\text{u}=\arctan\left(\text{u}\right)-\frac{\ln\left|1+\text{u}^2\right|}{2}+\text{C}_1\tag3$$

*For the RHS:
$$\int\frac{1}{x}\space\text{d}x=\ln\left|x\right|+\text{C}_2\tag4$$


So, we get:
$$\arctan\left(\text{r}\left(x\right)\right)-\frac{\ln\left|1+\text{r}\left(x\right)^2\right|}{2}=\ln\left|x\right|+\text{C}\tag5$$
Now, set $x\cdot\text{r}\left(x\right)=\text{y}\left(x\right)$ back:
$$\arctan\left(\frac{\text{y}\left(x\right)}{x}\right)-\frac{\ln\left|1+\left(\frac{\text{y}\left(x\right)}{x}\right)^2\right|}{2}=\ln\left|x\right|+\text{C}\tag6$$
Simplify a bit:
$$2\arctan\left(\frac{\text{y}\left(x\right)}{x}\right)=\ln\left(x^2+\text{y}\left(x\right)^2\right)+\text{C}\tag7$$
A: This is a dimensionally homogenous ODE. 
Let $u = \frac{y}{x}$
Then we have $\frac{du}{dx} = \frac{x\frac{dy}{dx} - y}{x^{2}}$
So $\frac{du}{dx} = \frac{\frac{dy}{dx} - u}{x}$
$\frac{dy}{dx} = x\frac{du}{dx} + u$
Substituting this in:
$x\frac{du}{dx} + u = \frac{x + ux}{x-ux}$
$x\frac{du}{dx} + u = \frac{1+u}{1-u}$
$x\frac{du}{dx} = \frac{1+u - u + u^{2}}{1-u}$
$x\frac{du}{dx} = \frac{u^{2} + 1}{1-u}$
$\int \frac{1-u}{u^{2}+1}du = \int \frac{1}{x} dx$
Can you solve from here?
