Question

Find the general solution of the given higher-order differential equation. d3u dt3 + d2u dt2 − 2u = 0

Answer 1

$\dfrac{\mathrm d^3u}{\mathrm dt^3}+\dfrac{\mathrm d^2u}{\mathrm dt^2}-2u=0$

This ODE has characteristic equation

$r^3+r^2-2=(r^3-r)+(r^2+r-2)=r(r^2-1)+(r+2)(r-1)$

$=(r(r+1)+(r+2))(r-1)=(r^2+2r+2)(r-1)=0$

which has roots at $r=1,-1\pm i$. Then the characteristic solution to the ODE is

$u(t)=C_1e^t+C_2e^{(-1+i)t}+C_3e^{(-1-i)t}$

$\implies\boxed{u(t)=C_1e^t+C_2e^{-t}\cos t+C_3e^{-t}\sin t}$

Answer 2

According to the roots of the characteristic equation, the solution is given by:

$u(t) = Ae^{-t} + e^{t}(B\cos{t} + C\sin{t})$

In which A, B and C are constants related to the initial conditions.

The differential equation is:

$\frac{d^3u}{du^3} + \frac{d^2u}{du^2} - 2u = 0$

The characteristic equation is:

$r^3 + r^2 - 2r = 0$

It's roots are:

$r_1 = 1, r_2 = -1 - i, r_3 = -1 + i$

• Relative to $r_1$, since it is real, we have $u_1 = Ae^{-t}$
• Relative to $r_2$ and $r_3$, since it is complex-conjugate, we have $u_2 = e^{t}(B\cos{t} + C\sin{t})$

Thus, the solution is:

$u(t) = u_1(t) + u_2(t)$

$u(t) = Ae^{-t} + e^{t}(B\cos{t} + C\sin{t})$

In which A, B and C are constants related to the initial conditions.

A similar problem is given at https://brainly.com/question/13244107