Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Theorem: Let $M(x,y)$ and $N(x,y)$ and $M_y(x,y)$, $N_x(x,y)$ be continuous functions in the rectangle $S:|x-x_0|<a$, $|y-y_0|<b$, where $(0<a, b<\infty)$. Then the differential equation $M +Ny' =0$ is exact. $M_x=M_y$


Problem: Now I have to show that the solution of $M+Ny'=0$ is given by $$\int_{x_0}^xM(s,y)ds + \int_{y_0}^y N(x_0,t)dt = c $$

My knowledge: Once the DE is exact, its implict solution is u(x,y) = c, where $M_y=N_x$ and $$u(x,y)=\int_{x_0}^x M(s,y)ds+g(y) $$. $g(y)$ is an arbitrary function of $y$ and plays the role of the constant of integration. I can obtain g by using the equation $u_y=N$: $$N(x,y)=\frac{\delta}{\delta y}\int_{x_0}^x M(s,y)ds+g' (y)$$. $$g(y)=\int_{y_0}^y N(x,t)dt - \int_{x_0}^{x} M(s,y)ds + \int_{x_0}^x M(s,y_0)ds+g(y_0)$$.

therefore $$u(x,y)=\int_{y_0}^y N(x,t)dt + \int_{x_0}^x M(s,y_0)ds =c$$

share|cite|improve this question
I think something like $M_y=N_x$ should be added to the body of theorem. In fact you are verifing the sufficient condition. – Babak S. Nov 14 '12 at 14:03
"$b>\infty$"? ${{}}$ – Michael Hardy Nov 14 '12 at 14:31
ty, I fixed that – Applied mathematician Nov 14 '12 at 14:48

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.