Solving Non-Exact Differential Equations with Integrating Factor

[mr_coffee]

Hello everyone I understand how to solve exact equations, but what happens when they arnt' exact? I'm confused on what I'm suppose to do! Does anyone feel like explaning hte process to me, if given an integrating factor/> or give me a website? Here is my problem:
Check that the equation below is not exact but becomes exact when multiplied by the integrating factor.
http://cwcsrv11.cwc.psu.edu/webwork2_files/tmp/equations/dd/9b5141797d9106d32db40a0eac2dc81.png
Integrating Factor:
http://cwcsrv11.cwc.psu.edu/webwork2_files/tmp/equations/26/0eab7b29675ba3a7a3db0668549eb81.png
Solve the differential equation.
You can define the solution curve implicitly by a function in the form
http://cwcsrv11.cwc.psu.edu/webwork2_files/tmp/equations/6b/90f9e4bd0e95299118e1c30fd8981d1.png
http://cwcsrv11.cwc.psu.edu/webwork2_files/tmp/equations/8d/8616ca1dd6ce230911485b98a57fa31.png ?


Thank you!

 

[HallsofIvy]

Do you understand what an "integrating factor" is?

If a differential equation is not exact then there always exists some function f(x,y) so that multiplying the equation by it makes it exact. Unfortunately, it's most often very difficult (if not impossible) to find that integrating factor!

In this case your equation is x²y³dx+ x(1+ y²)dy= 0. Yes, that is NOT exact because (x²y³)<sub>y= 3x<sup>2y2 which is not the same as (x(1+ y2))x= 1+y2.

Fortunately, you were told that $\frac{1}{xy^3}$ is an integrating factor. That means that multiplying the equation by that:
(1/xy2){x2y3dx+ x(1+y2)dy
= ydx+ (1+y2)y2)dy= 0.

Is that exact? It certainly should be! If it is exact can you solve it?</sup></sub>

 

[mr_coffee]

OOoo! Awesome, thanks a lot IVEY as always! It worked great! I got:
http://cwcsrv11.cwc.psu.edu/webwork2_files/tmp/equations/3c/d002938fe55dd46091d4edaff05c161.png