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June 27th, 2018, 11:24 AM  #1 
Newbie Joined: Jun 2018 From: Netherlands Posts: 2 Thanks: 1  Differential equation problem
Hi guys, I have a (small) problem with a differential equation I want to solve. Now I know the answer, but I do not know how to get there, which is essential. The differential equation is the following: dm/dp = a+b*p+c*m The answer is: m=u*exp(c*p)(1/c)*(b*p+b/c+c) where u is the constant of integration. Can somebody please explain the steps to me? Many thanks in advance!! 
June 27th, 2018, 11:46 AM  #2 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 520 Math Focus: Calculus/ODEs 
We are given to solve: $\displaystyle \frac{dm}{dp}=a+bp+cm$ I would first arrange in standard linear form: $\displaystyle \frac{dm}{dp}cm=a+bp$ Compute the integrating factor: $\displaystyle \mu(p)=\exp\left(c\int \,dp\right)=e^{cp}$ Multiply the ODE in standard form by this factor: $\displaystyle e^{cp}\frac{dm}{dp}ce^{cp}m=(a+bp)e^{cp}$ Rewrite the LHS: $\displaystyle \frac{d}{dp}\left(e^{cp}m(p)\right)=(a+bp)e^{cp}$ Integrate w.r.t $\displaystyle p$: $\displaystyle e^{cp}m(p)= \frac{e^{cp}(ac+bcp+b)}{c^2}+C$ Hence: $\displaystyle m(p)=Ce^{cp}\frac{ac+bcp+b}{c^2}$ This is almost the result you posted as the solution...did you copy it correctly? 
June 27th, 2018, 11:57 AM  #3 
Newbie Joined: Jun 2018 From: Netherlands Posts: 2 Thanks: 1 
Thank you so much! You're right, I made a small typo in the answer. The last c should be an a.

June 27th, 2018, 01:02 PM  #4 
Global Moderator Joined: Dec 2006 Posts: 20,285 Thanks: 1968 
I'll assume that c is not zero. The equation implies e^(cp)dm/dp  cme^(cp) = ae^(cp) + bpe^(cp). Integrating w.r.t. p gives e^(cp)m = (a/c)e^(cp)  (b/c)pe^(cp)  (b/c²)e^(cp) + u, where u is a constant. Multiplying by e^(cp) gives m = ue^(cp)  (b/c)p  b/c²  a/c, which is slightly different from what you gave as the answer. If c is zero, what solution do you get? 

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