Differential equations - transforming from one to another

  • Thread starter Basheesh
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  • #1
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Homework Statement



[tex]
m\frac{d^{2}x}{dt^{2}}+c\frac{dx}{dt}+kx=F_{0}cos\omega t\\
LC\frac{d^{2}V_{c}}{dt^{2}}+RC\frac{dV_{c}}{dt}+V_{c}=V_{0}cos\omega t
[/tex]


These two equations are for two different physical phenomena. The first is for a mass moving along an x axis. The mass is affected by a springforce k, a dampener with constant c and an outer force F(t), here set to [tex]F_0cost(\omega t)[/tex].
Explanation of the second equation follows. The gist is that they are mathematically the same (I'm translating from danish, let me know if you need the rest).

a) show that both differential equations can be brought to the form:

1:
[tex]
\frac{dx^{2}}{dt^{2}}+2\alpha \frac{dx}{dt}+\omega_{0}^{2}x=fcos\omega t
[/tex]

Where [tex]\alpha \geq 0[/tex] and we assume [tex]\omega > 0[/tex]


Homework Equations



1:
[tex]
m\frac{d^{2}x}{dt^{2}}+c\frac{dx}{dt}+kx=F_{0}cos\omega t, LC\frac{d^{2}V_{c}}{dt^{2}}+RC\frac{dV_{c}}{dt}+V_{c}=V_{0}cos\omega t
[/tex]


2:
[tex]
\frac{dx^{2}}{dt^{2}}+2\alpha \frac{dx}{dt}+\omega_{0}^{2}x=fcos\omega t
[/tex]

The Attempt at a Solution



I'm not even sure where to start with this.
So I'm thinking that I somehow need to use:
[tex]
\omega_{0}^{2} = c^2 + 4mk
[/tex]

and possible take the integral of the whole thing. But I am pretty lost.
 
Last edited:

Answers and Replies

  • #2
311
1
which question are you having trouble? 1 or 2?
 
  • #3
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I'm having trouble with a) for now. I edited out b to avoid confusion. But I might add it later, as I think I might have trouble with that one too.
 
  • #4
hunt_mat
Homework Helper
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Use, [tex]x=Ae^{kt}+Be^{-kt}[/tex] to find the general solution and to find the particular integral [tex]x=C\cos\omega t+D\sin\omega t[/tex].
 
  • #5
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But how does that help me writing the two equations in the form of (1)?
 

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