Question

In Problems 15 to 28 find the general solution of the given higher-order differential equation.

$\frac{{\mathbf{d}}^{\mathbf{3}}\mathbf{u}}{\mathbf{d}{\mathbf{t}}^{\mathbf{3}}}\mathbf{+}\frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{u}}{\mathbf{d}{\mathbf{t}}^{\mathbf{2}}}\mathbf{-}\mathbf{2}\mathit{u}\mathbf{=}\mathbf{0}$

Short answer

$u=c_1{e}^t+e^{-t}\left[c_2\cos t+c_3\sin t\right]$

Step-by-step solution

Differentiate the equation

Consider $y=e^{mx}$as the solution of the differential equation,

Substitute $y=e^{mx},y\text{'}=m{e}^{mx}$and $y\text{'}\text{'}=m^2{e}^{mx},y\text{'}\text{'}\text{'}=m^3{e}^{mx}$into $\frac{d^{3}u}{d{t}^3}+\frac{d^{2}u}{d{t}^2}-2u=0$to obtain the auxiliary equation.

Substitute in the equation

$$\begin{gathered} \frac{d^{3}u}{d{t}^3}+\frac{d^{2}u}{d{t}^2}-2u=0 \\ {m}^3{e}^{mt}+m^2{e}^{mt}-2e^{mt}=0 \\ {e}^{mt}\left(m^3+m^2-2\right)=0 \\ {e}^{mt}\left[(m-1)\left(m^2+2m+2\right)\right]=0 \end{gathered}$$

For any real $x,e^{mx}\ne 0$and m = 1.

From the equation $m^2+2m+2=0,$the value of a is 1, the value of b is , and the value of c is .

Substitute in the equation $m=\frac{-b\pm \sqrt{b^2-4ac}}{2a}.$

$$\begin{gathered} m=\frac{-\left(2\right)\pm \sqrt{{\left(2\right)}^2-4\left(1\right)\left(2\right)}}{2\left(1\right)} \\ =\frac{-2\pm \sqrt{4-8}}{2} \\ =\frac{-2\pm \sqrt{-4}}{2} \\ =\frac{-2\pm 2i}{2} \\ =-1\pm i \end{gathered}$$

Find the general solution of the Higher-order differential equation

Substitute $m_1=1,m_2=-1+i,m_3=-1-i,$in the general solution

$$\begin{gathered} u=c_1{e}^{m_{1}t}+c_2{e}^{m_{2}t}+c_3{e}^{m_{3}t} \\ =c_1{e}^{1·t}+c_2{e}^{(-1+i)t}+c_3{e}^{(-1-i)t} \\ =c_1{e}^t+c_2{e}^{-t}·e^{it}+c_3{e}^{-t}·e^{-it}\left[\because e^{i\theta }=\cos \theta +\sin \theta \right] \\ =c_1{e}^t+e^{-t}\left[c_2(\cos t+\sin t)+c_3(\cos t-\sin t)\right] \\ =c_1{e}^t+e^{-t}\left[c_2\cos t+c_2\sin t+c_3\cos t-c_3\sin t\right] \\ =c_1{e}^t+e^{-t}\left[\left(c_2+c_3\right)\cos t+\left(c_2-c_3\right)\sin t\right] \\ =c_1{e}^t+e^{-t}\left[c_2\cos t+c_3\sin t\right] \end{gathered}$$

Hence the general solution is$\mathit{u}\mathbf{=}{\mathit{c}}_{\mathbf{1}}{\mathit{e}}^{\mathbf{t}}\mathbf{+}{\mathit{e}}^{\mathbf{-}\mathbf{t}}\left[c_{2}cost+c_{3}sint\right]$