Solve this 3rd Order Differential Equation

[eric2921]

Homework Statement

d^3x/dt^3 - d^2x/dt^2 = 3e^t - sin(t)

Homework Equations

not sure. maybe sin(t)=.5*i*e^(-i*t)-.5*i*e^(i*t)

The Attempt at a Solution

I'm pretty sure we start by finding the complementary solution by setting the right side equal to 0. So we have:

D^3 x - D^2 x = 0
D^2 [D-1] x = 0

then we solve for D and get 0, 0 and 1 for solutions.
this gives us:

xcomp. = [c1+t c2] e^(0t) + c3 e^t
xcomp. = c1 + c2 t + c3 e^t

now its my understanding that i need to find particular solution because:

x = xcomp. + xpart.

So i was told that the solutions to the particular solution are 1, i and -i. I don't understand why that is though but if that's the case then i think:

xpart. = t e^t + A cos(t) + B cos(t)

and

x= c1 + c2 t + c3 e^t + t e^t + A cos(t) + B cos(t)

I'm stuck here, I'm pretty sure it's correct up to this point although i could be wrong. I don't know to find A or B though which i think all i have left to do. Thanks for any help!

 

[gomunkul51]

Hi,

your template is correct: xpart. = t e^t + A cos(t) + B sin(t). a little thing you forgot is a constant in front of t e^t --> K t e^t

the next step is to "plug in" only the xpart. solution into the original equation and find the constants K, A, B. then you are done.

P.S. this is the form of the full answer: x(t) = C1 + C2 t + C3 e^t + K t e^t + A cos(t) + B sin(t)
** you can find C1, C2, C3 if you have boundary conditions (starting values).

good luck.

 

[iamalexalright]

Why is xpart = t e^t + A cos(t) + B cos(t) ?

shouldnt it be:
xpart = t e^t + A cos(t) + B sin(t) or am I just missing something?

 

[hunt_mat]

You can integrate twice directly to reduce this third order down to a first order

 

[vela]

Homework Statement

d^3x/dt^3 - d^2x/dt^2 = 3e^t - sin(t)

Homework Equations

not sure. maybe sin(t)=.5*i*e^(-i*t)-.5*i*e^(i*t)

The Attempt at a Solution

I'm pretty sure we start by finding the complementary solution by setting the right side equal to 0. So we have:

D^3 x - D^2 x = 0
D^2 [D-1] x = 0

then we solve for D and get 0, 0 and 1 for solutions.
this gives us:

xcomp. = [c1+t c2] e^(0t) + c3 e^t
xcomp. = c1 + c2 t + c3 e^t

now its my understanding that i need to find particular solution because:

x = xcomp. + xpart.

Good so far.

So i was told that the solutions to the particular solution are 1, i and -i. I don't understand why that is though but if that's the case then i think:

xpart. = t e^t + A cos(t) + B cos(t)

and

x= c1 + c2 t + c3 e^t + t e^t + A cos(t) + B cos(t)

Almost.

The function sin t can be written in terms of complex exponentials:

[tex]\sin t = \frac{e^{it} - e^{-it}}{2i}[/tex]

so the righthand side can be written as

[tex]e^t - \frac{1}{2i} e^{it} + \frac{1}{2i} e^{-it}[/tex]

You should be able to see that

(D+i)(D-i)(D-1) [3 e^t - sin t] = 0

Those are the 1, i, and -i roots they're referring to. It follows then that the complete solution x(t) satisfies the homogeneous equation

(D+i)(D-i)(D-1) D²(D-1)x(t) = 0

If you write down the solution to this equation, you'd get

[tex]x(t) = c_1 + c_2 t + c_3 e^t + A t e^t + B e^{it} + C e^{-it}[/tex]

or equivalently

[tex]x(t) = c_1 + c_2 t + c_3 e^t + A t e^t + B \cos t + C \sin t[/tex]

The first three terms are the homogeneous solution you found earlier. The last three terms make up x_p(t). In practice, you don't go through this long procedure. You just look at the terms of the forcing function and deduce what terms will arise in the particular solution (taking into account what type of terms already appear in the complementary solution). But it's good to know why this works.

I'm stuck here, I'm pretty sure it's correct up to this point although i could be wrong. I don't know to find A or B though which i think all i have left to do. Thanks for any help!

Now you want to plug [itex]x_p(t) = A t e^t + B \cos t + C \sin t[/itex] into the original differential equation, then match coefficients for the various terms. From the three resulting equations, you can solve for A, B, and C.