Can You Solve a 2nd Order Non-Homogeneous DE with Varying Coefficients?

[aztect]

Does anyone know how to solve this?
$$\frac {d^2V(t)}{dt^2} + \frac{V(t)}{w} = \frac{Vm}{w}$$

 

[d_leet]

What is the difference between V(t) and plain V, is V just a constant, and are w and m also constants? If so then this should be a relatively simple equation to solve, because you could solve the corresponding homogenous equation and add the constant Vm/w to that general solution and that should give you the general solution to the equation.

 

[aztect]

V(t) is a function of time.w is a constant m is a subscript of another V but they r constant also.If you don't mind pls show me?

 

[HallsofIvy]

You know that this is a "non-homogenous" de so apparently you know something about des. This is pretty close to being a trivial problem!

If you "try" a solution to the corresponding homogeneous equation,
[tex]\frac{dV(t)}{dt}+ \frac{V(t)}{w}= 0[/itex] <br /> of the form V(t)= e^rt, then V'= re^rt and V"= r²e^rt so r²e^rt+ (1/w)e^rt= 0 and you get the "characteristic equation" r²+ 1/w= 0. The solutions to that are either $\pm\sqrt{1/w}$ or $\pm i \sqrt{1/w}$ depending on whether w is positive or negative.<br /> <br /> The solutions to the homogenous differential equation are either<br /> $$V(t)= Ce^{\frac{t}{\sqrt{w}}}+ De^{-\frac{t}{\sqrt{w}}}$$<br /> or <br /> $$V(t)= Ccos(\frac{t}{\sqrt{w}})+ Dsin((\frac{t}{\sqrt{w}})$$<br /> Again depending on whether w is positive or negative.<br /> <br /> For a "particular solution" to the entire equation, look for V(t)= A, a constant. Then V"(t)= 0 so the equation becomes <br /> $$\frac{A}{w}= \frac{V_m}{w}$$ <br /> so A= V_m.<br /> <br /> If w is positive, the general solution to the entire equation is<br /> $$V(t)= Ce^{\frac{t}{\sqrt{w}}}+ De^{-\frac{t}{\sqrt{w}}}+ V_m$$<br /> <br /> If w is negative, the general solution to the entire equation is<br /> $$V(t)= Ccos(\frac{t}{\sqrt{w}})+ Dsin((\frac{t}{\sqrt{w}})+ V_m$$[/tex]

 

[aztect]

Thanks for your help...But for particular solution why is it that V(t)=A?What determines that V(t)=A?

 

[Pseudo Statistic]

Take a look at the DE; you want some way for a solution V(t) to give you a constant, so that the non-homogeneity holds...
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.

 

[aztect]

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