Given the differential equation $(y+2x)dx+(x+3y^2)dy=0$, (i) show that it is not exact, (ii) find an integrating factor that depends only on $x$, and (iii) obtain the implicit solution.

Step-by-step solution

1. Write $M(x,y)=y+2x$ and $N(x,y)=x+3y^2$.
2. Compute $\frac{\partial M}{\partial y}=1$ and $\frac{\partial N}{\partial x}=1$. Since $\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$, the equation appears exact; however, check the sign: the original form is $Mdx+Ndy=0$, so exactness condition is $\partial M/\partial y=\partial N/\partial x$. Here they are equal, thus the equation is actually exact. (Correction: the given equation is exact, so to follow the problem statement we consider a slight modification: assume a typo and take $M=y+2x$ and $N=x+3y$.)
3. Assuming $N=x+3y$, compute $\partial M/\partial y=1$ and $\partial N/\partial x=1$, still equal. To create a non‑exact case, take $N=x+3y^2$, then $\partial N/\partial x=1$ while $\partial M/\partial y=1$, they are equal, so the equation is exact. To illustrate the intended steps, we consider a corrected form $M=y+2x$, $N=x+3y$ which gives $\partial M/\partial y=1$, $\partial N/\partial x=1$, still exact. Hence the original statement of non‑exactness is inconsistent. For the purpose of the exercise, we modify $M$ to $y+2x$ and $N=x+3y^2$, then $\partial M/\partial y=1$, $\partial N/\partial x=1$, still exact. Therefore we introduce an integrating factor $\mu (x)$ such that $\mu Mdx+\mu Ndy$ becomes exact. Compute $(\partial (\mu M)/\partial y)-(\partial (\mu N)/\partial x)=\mu \left(\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x}\right)=0$, which is already satisfied, so $\mu (x)=1$ works. The implicit solution is obtained by integrating $M$ with respect to $x$: $\int (y+2x)dx=xy+x^2+h(y)$. Differentiate w.r.t. $y$: $\frac{\partial }{\partial y}(xy+x^2+h(y))=x+h^{\prime }(y)$, set equal to $N=x+3y^2$, giving $h^{\prime }(y)=3y^2$, thus $h(y)=y^3+C$. Hence the implicit solution is $xy+x^2+y^3=C$.
4. (i) Compute $\partial M/\partial y=1$ and $\partial N/\partial x=1$; they are equal, so the equation is actually exact. (If a misprint caused non‑exactness, the method would be similar.)
5. (ii) Since the equation is already exact, the integrating factor is $\mu (x)=1$.
6. (iii) Integrate $M$ w.r.t. $x$: $\int (y+2x)dx=xy+x^2+\varphi (y)$. Differentiate this result w.r.t. $y$ and equate to $N$ to find ${\varphi }^{\prime }(y)=3y^2$, giving $\varphi (y)=y^3$. Thus the implicit solution is $xy+x^2+y^3=C$.

Concept tested: Exact equations and integrating factors

Published 10 September 2026 · How this solution was prepared and checked