When solving the differential equations which are reducible to exact differential equations using Inspection cases for example:

Solve: $2xy^2 + ye^xdx = e^xdy$

The integrating factor would $1/y^2$ so, you get:

$$\frac{(ye^xdx - e^xdy)}{y^2} + 2xdx = 0$$ or, $$d(\frac{e^x}{y}) + 2xdx = 0$$

After that, you just integrate to get the solution.

My question is why don't we check if $$d(\frac{e^x}{y}) + 2xdx = 0$$ is exact before integrating? Is it not possible to do so?

Because if it were say, homogenous differential equation, you would always first check if it was exact before integrating to get solution.

  • $\begingroup$ If you picked the right integrating factor, then the new equation is exact. So presumably something in the derivation of the integrating factor tells you that it will work out. $\endgroup$ – Ian Aug 30 '15 at 22:36
  • $\begingroup$ Is it possible to check to be sure that it will work out? $\endgroup$ – cpx Aug 31 '15 at 1:50
  • $\begingroup$ You basically already did: $d (e^x/y+x^2)=0$. The fact that you can write it this way is another way of saying the new equation is exact. $\endgroup$ – Ian Aug 31 '15 at 10:09
  • $\begingroup$ I think I see now. Thanks for clarification! $\endgroup$ – cpx Sep 1 '15 at 17:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.