Question

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

$\mathbf{16}\frac{{\mathbf{d}}^{\mathbf{4}}\mathbf{y}}{\mathbf{d}{\mathbf{x}}^{\mathbf{4}}}\mathbf{+}\mathbf{24}\frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{y}}{\mathbf{d}{\mathbf{x}}^{\mathbf{2}}}\mathbf{+}\mathbf{9}\mathit{y}\mathbf{=}\mathbf{0}$

Short answer

$y=h_1\cos \frac{\sqrt{3}}{2}x+h_2\sin \frac{\sqrt{3}}{2}x+h_{3}x\cos \frac{\sqrt{3}}{2}x+h_{4}x\sin \frac{\sqrt{3}}{2}x$

Step-by-step solution

Determine the general equation of the higher – order differential equation

We have the fourth order differential equation

$16y^{\left(4\right)}+24y^{\text{'}\text{'}}+9y=0$

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 $16y^{\left(4\right)}+24y^{\text{'}\text{'}}+9y=0$to obtain the auxiliary equation.

$$\begin{gathered} 16m^4{e}^{mx}+24m^2{e}^{mx}+9e^{mx}=0 \\ {e}^{mx}\left(16m^4+24m^2+9\right)=0 \end{gathered}$$

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

$$\begin{gathered} 16m^4+24m^2+9=0 \\ \left(4m^2+3\right)\left(4m^2+3\right)=0 \end{gathered}$$

Then the roots are $m_{1,2}=\sqrt{-\frac{3}{4}}=\pm \frac{\sqrt{3}}{2}i\text{and}{m}_{3,4}=\pm \frac{\sqrt{3}}{2}i$which are complex and repeated.

Finally, since we have repeated roots $m_{1,3}=\frac{\sqrt{3}}{2}i\text{and}{m}_{2,4}=-\frac{\sqrt{3}}{2}i$, then we need to have an additional linearly independent solution for each repeated root, then we obtain the general solution of our differential equation $16y^{\left(4\right)}+24y^{\text{'}\text{'}}+9y=0$ as

$$\begin{gathered} x=\left(c_1{e}^{\left(\frac{\sqrt{3}i}{2}\right)x}+c_2{e}^{\left(-\frac{\sqrt{3}i}{2}\right)x}\right)(n+sx) \\ =\left(c_1\left(\cos \frac{\sqrt{3}}{2}x+\sin \frac{\sqrt{3}}{2}x\right)+c_2\left(\cos \frac{\sqrt{3}}{2}x-\sin \frac{\sqrt{3}}{2}x\right)\right)(n+sx) \\ =\left(k_1\cos \frac{\sqrt{3}}{2}x+k_2\sin \frac{\sqrt{3}}{2}x\right)(n+sx) \\ =h_1\cos \frac{\sqrt{3}}{2}x+h_2\sin \frac{\sqrt{3}}{2}x+h_{3}x\cos \frac{\sqrt{3}}{2}x+h_{4}x\sin \frac{\sqrt{3}}{2}x \end{gathered}$$

Where we have $$\begin{gathered} h_1=n\times k_1=n\times \left(c_1+c_2\right) \\ {h}_2=n\times k_2=n\times \left(c_1-c_2\right) \\ {h}_3=s\times k_1=s\times \left(c_1+c_2\right) \\ {h}_4=s\times k_2=s\times \left(c_1-c_2\right) \end{gathered}$$ are arbitrary constants.

Hence, the general solution is

$\mathit{y}\mathbf{=}{\mathit{h}}_{\mathbf{1}}\mathit{c}\mathit{o}\mathit{s}\frac{\sqrt{\mathbf{3}}}{\mathbf{2}}\mathit{x}\mathbf{+}{\mathit{h}}_{\mathbf{2}}\mathit{s}\mathit{i}\mathit{n}\frac{\sqrt{\mathbf{3}}}{\mathbf{2}}\mathit{x}\mathbf{+}{\mathit{h}}_{\mathbf{3}}\mathit{x}\mathit{c}\mathit{o}\mathit{s}\frac{\sqrt{\mathbf{3}}}{\mathbf{2}}\mathit{x}\mathbf{+}{\mathit{h}}_{\mathbf{4}}\mathit{x}\mathit{s}\mathit{i}\mathit{n}\frac{\sqrt{\mathbf{3}}}{\mathbf{2}}\mathit{x}$