Integrating Factors for Solving Differential Equations

[kayella19]

solve n(9m²+18m-mn²-n²)dm + m(9m+9mn-3n²-n³)dn = 0

 

[Greg]

Any thoughts on how to begin?

 

[kayella19]

I have no any ideas. But in reference with Differential Equation 7th edition by rainville and bedient. It's under on ADDITIONAL TOPIC IN FIRST ORDER DIFF. EQUATION

 

[MarkFL]

I have no any ideas. But in reference with Differential Equation 7th edition by rainville and bedient. It's under on ADDITIONAL TOPIC IN FIRST ORDER DIFF. EQUATION

I haven't solved this ODE, but I think I would first check for exactness. Is this an exact ODE?

 

[Ackbach]

solve n(9m²+18m-mn²-n²)dm + m(9m+9mn-3n²-n³)dn = 0

I haven't solved this ODE, but I think I would first check for exactness. Is this an exact ODE?

This is an extremely annoying kind of differential equation, where you need an integrating factor not of the usual form. I have to admit that I cheated by using Wolfram|Alpha, which can find the solution. Here's a form of integrating factor that works:

$$\mu=e^{r\cdot n+s\cdot m},$$

with $r$ and $s$ constants to be determined.

 

[MarkFL]

This is an extremely annoying kind of differential equation, where you need an integrating factor not of the usual form. I have to admit that I cheated by using Wolfram|Alpha, which can find the solution. Here's a form of integrating factor that works:

$$\mu=e^{r\cdot n+s\cdot m},$$

with $r$ and $s$ constants to be determined.

Yes, I cannot find an integrating factor using the methods with which I am familiar. I also initially used W|A to find the solution, and at least you were able to determine the form of an integrating factor that works. (Yes)

 

[MarkFL]

Note: This was done by taking the solution given by W|A and essentially "working backwards."

We are given to solve:

$$n\left(9m^2+18m-n^2-mn^2\right)dm+m\left(9m+9mn-3n^2-n^3\right)dn=0$$

Arrange as:

$$\left(n(9m-n^2)+9mn+mn(9m-n^2)\right)dm+\left(m(9m-n^2)-2mn^2+mn(9m-n^2)\right)dn=0$$

Further arrange as:

$$dm\cdot n(9m-n^2)dm+m(9m-n^2)dn+mn\left(9dm-2ndn\right)+mn(9m-n^2)\left(dm+dn\right)=0$$

Looking at the last factor in the last rem on the LHS, we observe that an integrating factor is $\mu=e^{m+n}$:

$$dm\cdot n(9m-n^2)e^{m+n}+m\cdot dn\cdot(9m-n^2)e^{m+n}+mn\left(9\,dm-2n\,dn\right)e^{m+n}+mn(9m-n^2)e^{m+n}\left(dm+dn\right)=0$$

Or:

$$n(9m-n^2)e^{m+n}+m\d{n}{m}\cdot(9m-n^2)e^{m+n}+mn\left(9-2n\d{n}{m}\right)e^{m+n}+mn(9m-n^2)e^{m+n}\left(1+\d{n}{m}\right)=0$$

This can be written as:

$$\frac{d}{dm}\left(mn(9m-n^2)e^{m+n}\right)=0$$

Integrate w.r.t $m$:

$$\int \frac{d}{dm}\left(mn(9m-n^2)e^{m+n}\right)\,dm=0\int \,dm$$

And so the solution is given implicitly by:

$$mn(9m-n^2)e^{m+n}=c_1$$

 

[DrWahoo]

Nice write up MarkFL. I konw Dr.Agarwal's book here gives some clues on the integrating factor for this particular problem. An Introduction to Ordinary Differential Equations | Ravi P. Agarwal | Springer He actually offers several generic integrating factors for different ODE's and this is one of the forms he provides with a very arbitrary proof.

I have worked with him on a couple research projects, and he offers a very good proof showing the first step solving by induction. He doesn't clearly state the n+1 case, but it does in fact admit an integrating factor of the form given above confirming the above calculations.