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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

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1 Answer 1

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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 $3xy+x^2y'+y^2+xyy'=(3xy+y^2)+(x^2+xy)y'=0$.

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