0
$\begingroup$

I have a problem where:

$\ddddot{x} - 2 \ddot{x} + x = 0$

With boundary conditions

$x(0) = 1, \dot{x}(0) = 2, x(\infty) = 0, \dot{x}(\infty) = 0$

So I get my characteristic equation:

$s^4 -2s^2 + 1 = 0$

$s = 1,1,-1,-1$

I have to add a couple extra t terms to my general solution due to the repeated roots:

$x(t) = A_1e^t + A_2te^t + A_3e^{-t} + A_4te^{-t}$

To setup my simultaneous equations to solve for the coefficients, I need to get the derivative:

$\dot{x}(t) = A_1e^t + A_2(e^t + te^t) - A_3e^{-t} + A_4(e^{-t} - te^{-t})$

So I can quickly setup my equations using t=0:

$x(0) = 1 = A_1 + A_3$
$\dot{x}(0) = 2 = A_1 + A_2 - A_3 + A_4$

but for the infinity terms, I'm getting equations with infinity in them...not sure what to do. Any suggestions?

$\endgroup$
1
  • 1
    $\begingroup$ In order for the solution to be bounded, I think A1 and A2 has to be zero. If they are nonzero, then x(infinity) will not be zero. For the derivative, A1 and A2 is also 0. Then you obtain the simultaneous questions with A1 and A2 eliminated. Two equations, two unknowns. A3=1, A4=3, A1=A2=0 $\endgroup$
    – Novice
    Sep 27 '14 at 17:34
3
$\begingroup$

Note that $t\mathrm{e}^t$ is asymptotically larger than the other solutions, and tends to $\infty$ as $t$ tends to $\infty$. Hence $A_2=0$. Similarly $A_1=0$, as $\mathrm{e}^t\to\infty$, while the rest of the terms tend to 0.

Hence $$ x(t)=(at+b)\mathrm{e}^{-t}. $$ The constants $a,b$ can be found from the initial data.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.