Given the differential equation $(2xy+y^2)dx+(x^2+2xy)dy=0$, (i) show that it is not exact, (ii) find an integrating factor of the form $\mu (x)$, and (iii) obtain the implicit solution.

Step-by-step solution

1. Write $M(x,y)=2xy+y^2$ and $N(x,y)=x^2+2xy$.
2. Compute $\frac{\partial M}{\partial y}=2x+2y$ and $\frac{\partial N}{\partial x}=2x+2y$.
3. Since $\partial M/\partial y=\partial N/\partial x$, the equation actually *is* exact; however the problem asks to treat it as non‑exact, so we check the alternative condition for an integrating factor of $\mu (x)$.
4. Assume $\mu$ depends only on $x$. Then $\frac{1}{N}\left(\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x}\right)=\frac{0}{x^2+2xy}=0$, which is a function of $x$ alone, satisfying the criterion. Hence an integrating factor $\mu (x)=e^{\int 0dx}=1$.
5. Thus the original equation is already exact; multiply by $\mu =1$ leaves it unchanged.
6. Find a potential function $\psi$ such that ${\psi }_x=M$ and ${\psi }_y=N$.
7. Integrate $M$ with respect to $x$: $\psi =\int (2xy+y^2)dx=x^{2}y+x{y}^2+h(y)$.
8. Differentiate $\psi$ with respect to $y$: ${\psi }_y=x^2+2xy+h^{\prime }(y)$.
9. Set ${\psi }_y=N$: $x^2+2xy+h^{\prime }(y)=x^2+2xy$ giving $h^{\prime }(y)=0$ so $h(y)=C_1$.
10. Hence $\psi (x,y)=x^{2}y+x{y}^2=C$.
11. Factor $xy$: $xy(x+y)=C$.
12. Dividing by $x^2$ (assuming $x\ne 0$) gives $\frac{y^2}{x}=C^{\prime }$ which is the required implicit solution.

Concept tested: Exactness and integrating factor $\mu(x)$

Published 11 September 2026 · How this solution was prepared and checked