2nd Order De Solution

  • Thread starter Peregrine
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  • #1
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Main Question or Discussion Point

I am familiar with how to solve a second order, non-homogenous DE with constants, i.e.

[tex]\frac {\partial^2X(t)}{\partial t^2} + \frac{\partial X(t)}{\partial t} = C[/tex]

by first solving the homogenous eqn, then setting the equation equal to a constant, yielding a sol'n of

[tex]X(t)= Ae^{0}+ Be^{-t}+ C[/tex]

But how does one solve a 2nd order equation that also has another t variable in it, such as:

[tex]\frac {\partial^2X(t)}{\partial t^2} + \frac{1}{t} \frac{\partial X(t)}{\partial t} = C[/tex]?
 

Answers and Replies

  • #2
First of all, you only seem to have one independent variable, so it may suitable to express your equation as

[tex]\frac{d^2 X(t)}{dt^2} + \frac{1}{t} \frac{d X(t)}{dt} = C[/tex]

(note total derivative, not partial). Also, since no X(t) appears outside a derivative, you effectively have a first order equation, namely

[tex]\frac{dp}{dt} + \frac{p}{t} = C [/tex]

where

[tex]p(t) = \frac{dX(t)}{dt} [/tex]

Now, any first order equation of the form

[tex]\frac{dy}{dx} + s(x) y + r(x) = 0 [/tex]

has the solution

[tex]y(x) = -e^{-\int s(x) dx} \int r(x) e^{\int s(x) dx} dx [/tex]

(just differentiate this and you'll see it works) Hence you can solve for p(t), and then for X(t).
 
Last edited:
  • #3
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Ah, that's a very nice way of framing the equation, I hadn't thought of that. Thanks!
 

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