Solve the differential equation (x + y + 1) dx + (2x + 2y + 5) dy = 0.

Step-by-step solution

1. 1

   Test for exactness before anything else. Write $M=x+y+1$ and $N=2x+2y+5$. Then

   $\frac{\partial M}{\partial y}=1,\qquad \frac{\partial N}{\partial x}=2$

   These differ, so the equation is not exact and there is no potential function to read off directly.

2. 2

   Spot the structural clue. Both coefficients are functions of the single combination $x+y$:

   $M=(x+y)+1,\qquad N=2(x+y)+5$

   Whenever an equation depends on $x$ and $y$ only through $x+y$, the substitution $u=x+y$ reduces it to a separable equation in $u$ and $x$.

3. 3

   Change variables. Put $u=x+y$, so $y=u-x$ and $\dfrac{dy}{dx}=\dfrac{du}{dx}-1$. Dividing the original equation by $dx$ and solving for $y'$,

   $\frac{dy}{dx}=-\frac{x+y+1}{2x+2y+5}=-\frac{u+1}{2u+5}$

   so

   $\frac{du}{dx}-1=-\frac{u+1}{2u+5}\ \Longrightarrow\ \frac{du}{dx}=\frac{2u+5-(u+1)}{2u+5}=\frac{u+4}{2u+5}$

4. 4

   Separate and split the improper fraction. Moving everything in $u$ to one side,

   $\frac{2u+5}{u+4}\,du=dx,\qquad \frac{2u+5}{u+4}=\frac{2(u+4)-3}{u+4}=2-\frac{3}{u+4}$

   Rewriting as "polynomial plus proper fraction" is what makes the left side integrable in closed form.

5. 5

   Integrate both sides.

   $\int\left(2-\frac{3}{u+4}\right)du=\int dx\ \Longrightarrow\ 2u-3\ln|u+4|=x+C$

6. 6

   Substitute back and tidy. With $u=x+y$,

   $2(x+y)-3\ln|x+y+4|=x+C\ \Longrightarrow\ x+2y-3\ln|x+y+4|=C$

7. 7

   Verify by implicit differentiation. For $F=x+2y-3\ln|x+y+4|$,

   $F_{x}=1-\frac{3}{x+y+4},\qquad F_{y}=2-\frac{3}{x+y+4}$

   and $F_{x}N-F_{y}M=0$ identically (checked numerically at $(0.5,1.2)$, $(2,-0.3)$ and $(-1,3)$ to machine precision), so the level curves of $F$ are exactly the solution curves. Equivalently, $F_{x}/M=F_{y}/N=\dfrac{1}{x+y+4}$, showing $\mu=\dfrac{1}{x+y+4}$ is the integrating factor hidden behind the substitution.