Problem with 2nd order linear differential equation

RolandIRL

hi all,
i'm doing a bit of applied maths study for exams, and i'm having trouble with 2nd order linear diff. equations. on the prob sheet given to us, i've solved the equation but the answer doesn't seem right.

Equation: d^2 x/dt^2 + x = cos(t)
initial conditions: x(0) = 0 and dx(0)/dt = 0

letting the LHS equal 0, i'm getting a characteristic equation by letting x = e^rt
(r^2 + 1)x=0
r = +i or r = -i

x= C cos(t) + D sin(t), for arbitrary constants C,D
this is the homogeneous solution

Particular solution: f(t) = cos(t)
x=E cos(t) + F sin(t)

the general solution is got by adding the homogeneous and particular solutions.
so...

x(t) = (C+E)cos(t) + (D+F)sin(t)

subject to the initial conditions, i'm getting that x(t) = 0, but i don't think this is right.
can someone point out where i'm going wrong?

thanks
whiteman19

Your particular solution is wrong. It is supposed to be any solution to your original equation d2x/dt^2 + x = cos t
So firstly, you wouldn't need to say E cos(t) + F sin(t) because you only need one solution. Secondly, watch what happens when you integrate this. Let's call it xp(t).

xp'(t) = -E sin (t) + F cos (t)
xp''(t) = -E cos (t) - F sin (t)

So xp''(t) + xp(t) = 0, while we want it to be cos(t).
Hope this helps.
Conor.

RolandIRL

thanks, i think i figured it out.

x'' + x = cos(t)

Homogeneous: x(h) = C cos(t) + D sin(t)

Particular: x(p) = t/2 sin(t)

x(t) = x(h) + x(h)
= C cos(t) + D sin(t) + t/2 sin(t)

x(0) = 0 C=0

x'(0) = 0 D= 0

x(t) = t/2 sin(t)

in relation to simple harmonic motion, this is an example of resonance, isn't it? where the amplitude is increasing as t tends to infinity.

Yes, the solution is right, and it is an example of resonance. In fact, the cos (t) in the inhomogeneous function would be an external force. Notice that the external force has the same period as your function.