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

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

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