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Does anyone know how to solve this?

[tex]\frac {d^2V(t)}{dt^2} + \frac{V(t)}{w} = \frac{Vm}{w}[/tex]

[tex]\frac {d^2V(t)}{dt^2} + \frac{V(t)}{w} = \frac{Vm}{w}[/tex]

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- Thread starter aztect
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- #1

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Does anyone know how to solve this?

[tex]\frac {d^2V(t)}{dt^2} + \frac{V(t)}{w} = \frac{Vm}{w}[/tex]

[tex]\frac {d^2V(t)}{dt^2} + \frac{V(t)}{w} = \frac{Vm}{w}[/tex]

- #2

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- #4

HallsofIvy

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Homework Helper

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If you "try" a solution to the corresponding homogeneous equation,

[tex]\frac{dV(t)}{dt}+ \frac{V(t)}{w}= 0[/itex]

of the form V(t)= e

The solutions to the homogenous differential equation are either

[tex]V(t)= Ce^{\frac{t}{\sqrt{w}}}+ De^{-\frac{t}{\sqrt{w}}}[/tex]

or

[tex]V(t)= Ccos(\frac{t}{\sqrt{w}})+ Dsin((\frac{t}{\sqrt{w}})[/tex]

Again depending on whether w is positive or negative.

For a "particular solution" to the entire equation, look for V(t)= A, a constant. Then V"(t)= 0 so the equation becomes

[tex]\frac{A}{w}= \frac{V_m}{w}[/tex]

so A= V_m.

If w is positive, the general solution to the entire equation is

[tex]V(t)= Ce^{\frac{t}{\sqrt{w}}}+ De^{-\frac{t}{\sqrt{w}}}+ V_m[/tex]

If w is negative, the general solution to the entire equation is

[tex]V(t)= Ccos(\frac{t}{\sqrt{w}})+ Dsin((\frac{t}{\sqrt{w}})+ V_m[/tex]

- #5

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- #6

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In your example, you want the differential equation just to give you a constant, right? So assuming V(t) = A and equating both sides would give you that constant.

Pretty hard to explain, but let's say you assuming V(t) = at + b; we have:

0 + (at+b)/w = (Vm/w)

at + b = Vm

And, writing it out in a slightly different way:

at + b = Vm + 0t

Equating the co-efficients, you have a = 0, Vm = b.

And that's another way to find the solution.

But what determines V(t) = A is because you have a V(t) term on the left side and you have a constant on the right side--- and since V'(t) = 0 and V''(t) = 0 for any constant A, V(t) = A would give you the constant as you want it on the right side.

Hope that made SOME sense. :P I tried.

- #7

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Thanks for all the help...This is a really good forum

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