I have no idea how to do this... and given y(1)=1
(xy+y2)*dy/dx=y2+xy+x2
Hello, fiksi!
It's homogeneous . . .$\displaystyle (xy+y^2)\,\frac{dy}{dx} \:=\:y^2+xy+x^2,\quad y(1) = 1$
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.
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