1. Simple DE

(This problem is from the chapter Substitution methods and exact equations.)

Find the general solution of $\displaystyle (x+y)y'=x-y$.

2. Re: Simple DE

Ok, let me give it a shot.

Let $\displaystyle x+y=v$.

$\displaystyle 1+\frac{dy}{dx}=\frac{dv}{dx}$

$\displaystyle \frac{dy}{dx}=\frac{dv}{dx}-1$

Substituting, we get

$\displaystyle v\left(\frac{dv}{dx}-1\right)=x-(v-x)=2x-v$

$\displaystyle v\frac{dv}{dx}=2x$

$\displaystyle \int vdv=\int 2xdx$

$\displaystyle \frac{v^2}{2}=x^2+K$

$\displaystyle \frac{(x+y)^2}{2}=x^2+K$

$\displaystyle x^2+2xy+y^2=2x^2+2K$

$\displaystyle x^2-2xy-y^2=-2K$

$\displaystyle x^2-2xy-y^2=C$, where $\displaystyle C=-2K$.

3. Re: Simple DE

Originally Posted by alexmahone
Ok, let me give it a shot.

Let $\displaystyle x+y=v$.

$\displaystyle 1+\frac{dy}{dx}=\frac{dv}{dx}$

$\displaystyle \frac{dy}{dx}=\frac{dv}{dx}-1$

Substituting, we get

$\displaystyle v\left(\frac{dv}{dx}-1\right)=x-(v-x)=2x-v$

$\displaystyle v\frac{dv}{dx}=2x$

$\displaystyle \int vdv=\int 2xdx$

$\displaystyle \frac{v^2}{2}=x^2+K$

$\displaystyle \frac{(x+y)^2}{2}=x^2+K$

$\displaystyle x^2+2xy+y^2=2x^2+2K$

$\displaystyle x^2-2xy-y^2=-2K$

$\displaystyle x^2-2xy-y^2=C$, where $\displaystyle C=-2K$.
Hi alexmahone,

Your method is correct. Alternatively you could use the substitution, $\displaystyle u=\frac{y}{x}$ as this is a differential equation of the "Homogeneous type".