Question

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

Answer 1

`(\mathrm d^3u)/(\mathrm dt^3)+(\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}`