I have the following differential equation:
$$(3xy+y^2)+(x^2+xy)y' = 0$$
which has the solution given implictly by:
$$x^3y+\frac{1}{2}x^2y^2 = c$$
I know that I can solve for $y$ and plug back into the differential equaiton, but this is not always the case. How to verify that this implicit solution is indeed a solution? I tried differentiating with respect to $x$ but I won't go back to that
Best Answer
Differentiating by $x^3y+{1\over 2}x^2y^2=c$ by $x$ you have $3x^2y+x^3y'+xy^2+x^2yy'=0=x(3xy+x^2y'+y^2+xyy')=0$ you deduce that $3xy+x^2y'+y^2+xyy'=(3xy+y^2)+(x^2+xy)y'=0$.