Solve the differential equation (e to the y, plus 6x) dx plus x e to the y dy equals 0.

Step-by-step solution

1. 1

   Name the two coefficient functions and test for exactness. With

   $M(x,y)=e^y+6x,\qquad N(x,y)=xe^y$

   the equation is exact if $\dfrac{\partial M}{\partial y}=\dfrac{\partial N}{\partial x}$. Here

   $\frac{\partial M}{\partial y}=e^y,\qquad \frac{\partial N}{\partial x}=e^y$

   They agree, so a potential function $F(x,y)$ with $F_x=M$ and $F_y=N$ exists — and no integrating factor is needed.

2. 2

   Integrate $M$ with respect to $x$ to start building $F$. Treating $y$ as a constant,

   $F(x,y)=\int\left(e^y+6x\right)dx=xe^y+3x^2+g(y)$

   The "constant" of integration is an unknown function of $y$, since anything depending only on $y$ vanishes under $\partial/\partial x$.

3. 3

   Differentiate the candidate with respect to $y$ and match $N$.

   $\frac{\partial F}{\partial y}=xe^y+g'(y)\stackrel{!}{=}N=xe^y$

4. 4

   Solve for the unknown function. Cancelling $xe^y$ from both sides leaves

   $g'(y)=0\quad\Rightarrow\quad g(y)=\text{constant}$

   That the $y$-dependence cancels completely is the exactness condition paying off; if a leftover $y$-only term had appeared, it would simply be integrated here.

5. 5

   Write the general solution implicitly. Exact equations have solutions of the form $F(x,y)=C$:

   $xe^y+3x^2=C$

   The arbitrary constant absorbs $g$'s constant.

6. 6

   Verify by differentiating the answer. Taking $d$ of both sides of $xe^y+3x^2=C$:

   $\left(e^y+6x\right)dx+xe^y\,dy=0$

   which is exactly the original equation ✓. (Solving explicitly gives $y=\ln\frac{C-3x^2}{x}$ where that argument is positive.)