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June 27th, 2018, 10:24 AM   #1
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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:


where u is the constant of integration. Can somebody please explain the steps to me? Many thanks in advance!!
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June 27th, 2018, 10:46 AM   #2
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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 $\displaystyle p$:

$\displaystyle e^{-cp}m(p)= -\frac{e^{-cp}(ac+bcp+b)}{c^2}+C$


$\displaystyle m(p)=Ce^{cp}-\frac{ac+bcp+b}{c^2}$

This is almost the result you posted as the solution...did you copy it correctly?
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June 27th, 2018, 10:57 AM   #3
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Thank you so much! You're right, I made a small typo in the answer. The last c should be an a.
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June 27th, 2018, 12:02 PM   #4
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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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