Question

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

$\frac{{\mathbf{d}}^{\mathbf{4}}\mathbf{y}}{\mathbf{d}{\mathbf{x}}^{\mathbf{4}}}\mathbf{-}\mathbf{7}\frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{y}}{\mathbf{d}{\mathbf{x}}^{\mathbf{2}}}\mathbf{-}\mathbf{18}\mathit{y}\mathbf{=}\mathbf{0}$

Short answer

$y=k_1{e}^{3x}+k_2{e}^{-3x}+k_3\cos \sqrt{2}x+k_4\sin \sqrt{2}x$

Step-by-step solution

Determine the general equation of the higher – order differential equation

We have the fourth order differential equation

$\frac{d^{4}y}{d{x}^4}-7\frac{d^{2}y}{d{x}^2}-18y=0$

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

Substitute$y=e^{mx},y\text{'}=m{e}^{mx}$and$y\text{'}\text{'}=m^2{e}^{mx},y\text{'}\text{'}\text{'}=m^3{e}^{mx}\text{and}{y}^{\left(4\right)}=m^4{e}^{mx}$into $y^{\left(4\right)}-2y^{\text{'}\text{'}}+y=0$ to obtain the auxiliary equation.

$$\begin{gathered} m^4{e}^{mx}-7m^2{e}^{mx}-18e^{mx}=0 \\ {e}^{mx}\left(m^4-7m^2-18\right)=0 \end{gathered}$$

For any real $x,e^{mx}\ne 0$, we have

$$\begin{gathered} m^4-7m^2-18=0 \\ (m-3)\left(m^3+3m^2+2m+6\right)=0 \\ (m-3)(m+3)\left(m^2+2\right)=0 \end{gathered}$$

Then the roots are $m_1=3,m_2=-3\text{and}{m}_{3,4}=\sqrt{-2}=\pm \sqrt{2}i$which are mix of real and complex.

Then the general solution of this homogeneous equation $\frac{d^{4}y}{d{x}^4}-7\frac{d^{2}y}{d{x}^2}-18y=0$ can be given as $$\begin{gathered} x=c_1{e}^{3x}+c_2{e}^{-3x}+c_3{e}^{\left(\sqrt{7}i\right)x}+c_3{e}^{\left(-\frac{\sqrt{7}i}{2}\right)t} \\ =c_1{e}^{3x}+c_2{e}^{-3x}+c_3(\cos \sqrt{2}x+\sin \sqrt{2}ix)+c_4(\cos \sqrt{2}x-\sin \sqrt{2}x) \\ =k_1{e}^{3x}+k_2{e}^{-3x}+k_3\cos \sqrt{2}x+k_4\sin \sqrt{2}x \end{gathered}$$where $$\begin{gathered} k_1=c_1 \\ {k}_2=c_2 \\ {k}_3=c_3+c_4 \\ {k}_4=c_3-c_4 \end{gathered}$$are arbitrary constants.

Hence, the general solution is$\mathit{y}\mathbf{=}{\mathit{k}}_{\mathbf{1}}{\mathit{e}}^{\mathbf{3}\mathbf{x}}\mathbf{+}{\mathit{k}}_{\mathbf{2}}{\mathit{e}}^{\mathbf{-}\mathbf{3}\mathbf{x}}\mathbf{+}{\mathit{k}}_{\mathbf{3}}\mathit{c}\mathit{o}\mathit{s}\sqrt{\mathbf{2}}\mathit{x}\mathbf{+}{\mathit{k}}_{\mathbf{4}}\mathit{s}\mathit{i}\mathit{n}\sqrt{\mathbf{2}}\mathit{x}$