# strange diff eqn

• Mar 3rd 2009, 01:01 PM
fiksi
strange diff eqn
I have no idea how to do this... and given y(1)=1

(xy+y2)*dy/dx=y2+xy+x2
• Mar 3rd 2009, 02:45 PM
Soroban
Hello, fiksi!

Quote:

$\displaystyle (xy+y^2)\,\frac{dy}{dx} \:=\:y^2+xy+x^2,\quad y(1) = 1$
It's homogeneous . . .

We have: . $\displaystyle \frac{dy}{dx} \;=\;\frac{y^2+xy + x^2}{xy + y^2}$

Divide top and bottom by $\displaystyle x^2\!:\;\;\frac{dy}{dx} \;=\; \frac{\left(\frac{y}{x}\right)^2 +\frac{y}{x} + 1} { \frac{y}{x} + \left(\frac{y}{x}\right)^2}$

Let: $\displaystyle v \,=\,\frac{y}{x} \quad\Rightarrow\quad y \,=\,vx \quad\Rightarrow\quad \frac{dy}{dx}\:=\:v + x\,\frac{dv}{dx}$

Substitute: .$\displaystyle v + x\,\frac{dv}{dx} \;=\;\frac{v^2+v+1}{v+v^2}$

. . . . $\displaystyle x\,\frac{dv}{dx} \:=\:\frac{v^2+v+1}{v+v^2} - v \quad\Rightarrow\quad x\,\frac{dv}{dx}\;=\;\frac{1+v-v^3}{v+v^2}$

Separate variables: . $\displaystyle \frac{v+v^2}{1+v-v^3}\,dv \;=\;\frac{dx}{x}$

Now "all we have to do" is integrate: .$\displaystyle \int\frac{v+v^2}{1+v-v^3}\,dv \;=\;\int\frac{dx}{x}$

. . back-substitute, and use that initial value.

But I'm having difficulty with that v-integral.
.
• Mar 4th 2009, 04:19 AM
fiksi
Quote:

Originally Posted by Soroban
Hello, fiksi!

But I'm having difficulty with that v-integral.
.

Thx very much... and I know why integral is impossible to do- they said today we should change in LHS in bracket x2 instead of y2. Thus we have (xy+x2) times all that etc. I guess it would be possible to do it now...