Simple DE

**alexmahone** · September 10th 2011, 04:57 PM

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

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

**alexmahone** · September 10th 2011, 08:07 PM

Ok, let me give it a shot.

Let $x+y=v$ .

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

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

Substituting, we get

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

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

$\int vdv=\int 2xdx$

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

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

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

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

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

**Sudharaka** · September 10th 2011, 10:37 PM

Originally Posted by alexmahone

Ok, let me give it a shot.

Let $x+y=v$ .

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

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

Substituting, we get

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

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

$\int vdv=\int 2xdx$

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

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

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

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

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

Hi alexmahone,

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

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