Question

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

$\frac{{\mathbf{d}}^{\mathbf{5}}\mathbf{u}}{\mathbf{d}{\mathbf{r}}^{\mathbf{5}}}\mathbf{+}\mathbf{5}\frac{{\mathbf{d}}^{\mathbf{4}}\mathbf{u}}{\mathbf{d}{\mathbf{r}}^{\mathbf{4}}}\mathbf{-}\mathbf{2}\frac{{\mathbf{d}}^{\mathbf{3}}\mathbf{u}}{\mathbf{d}{\mathbf{r}}^{\mathbf{3}}}\mathbf{-}\mathbf{10}\frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{u}}{\mathbf{d}{\mathbf{r}}^{\mathbf{2}}}\mathbf{+}\frac{\mathbf{d}\mathbf{u}}{\mathbf{d}\mathbf{r}}\mathbf{+}\mathbf{5}\mathit{u}\mathbf{=}\mathbf{0}$

Short answer

$u=c_1{e}^{-5r}+c_2{e}^r+c_{3}r{e}^r+c_4{e}^{-r}+c_{5}r{e}^{-r}$

Step-by-step solution

Determine the general equation of the higher – order differential equation

We have the fourth order differential equation

$\frac{d^{5}u}{d{r}^5}+5\frac{d^{4}u}{d{r}^4}-2\frac{d^{3}u}{d{r}^3}-10\frac{d^{2}u}{d{r}^2}+\frac{du}{dr}+5u=0$

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

Substitute $y=e^{\lambda r},y\text{'}=\lambda e^{\lambda r}$and$y\text{'}\text{'}={\lambda }^2{e}^{\lambda r},y\text{'}\text{'}\text{'}={\lambda }^3{e}^{\lambda r},y^{\left(4\right)}={\lambda }^4{e}^{\lambda r}\text{and}{y}^{\left(5\right)}={\lambda }^5{e}^{\lambda r}$ into $y^{\left(4\right)}-2y^{\text{'}\text{'}}+y=0$ to obtain the auxiliary equation.

$$\begin{gathered} d{\lambda }^5{e}^{\lambda r}+5{\lambda }^4{e}^{\lambda r}-2{\lambda }^3{e}^{\lambda r}-10{\lambda }^2{e}^{\lambda r}+\lambda e^{\lambda r}+5e^{\lambda r}=0 \\ {e}^{\lambda r}\left({\lambda }^5+5{\lambda }^4-2{\lambda }^3-10{\lambda }^2+\lambda +5\right)=0 \end{gathered}$$

Since $e^{\lambda r}$ cannot be 0, then we have

$$\begin{gathered} {\lambda }^5+5{\lambda }^4-2{\lambda }^3-10{\lambda }^2+\lambda +5(\lambda +5)\left({\lambda }^4-2{\lambda }^2+1\right)=0 \\ (\lambda +5)(\lambda +5){\left({\lambda }^2-1\right)}^2=0 \end{gathered}$$

Then the roots are ${\lambda }_1=-5,{\lambda }_{2,3}=1\text{and}{\lambda }_{4,5}=-1$ which are real and repeated.

Finally, since we have repeated roots, then we need to have an additional linearly independent solution for each repeated root, then we obtain the general solution of our

Differential equation $\frac{d^{5}u}{d{r}^5}+5\frac{d^{4}u}{d{r}^4}-2\frac{d^{3}u}{d{r}^3}-10\frac{d^{2}u}{d{r}^2}+\frac{du}{dr}+5u=0$as

$$\begin{gathered} =c_1{e}^{{\lambda }_{1}r}+n_1{e}^{{\lambda }_{2}r}\left(k_1+s_{1}x\right)+n_2{e}^{{\lambda }_{4}r}\left(k_2+s_{2}x\right) \\ =c_1{e}^{-5r}+n_1{e}^r\left(k_1+s_{1}x\right)+n_2{e}^{-r}\left(k_2+s_{2}x\right) \\ =c_1{e}^{-5r}+c_2{e}^r+c_{3}r{e}^r+c_4{e}^{-r}+c_{5}r{e}^{-r} \end{gathered}$$where we have$$\begin{gathered} c_2=n_1\times k_1 \\ {c}_3=n_1\times s_1 \\ {c}_4=n_2\times k_2 \\ {c}_5=n_2\times s_2 \end{gathered}$$ are arbitrary constants.

Hence the general solution is $\mathit{u}\mathbf{=}{\mathit{c}}_{\mathbf{1}}{\mathit{e}}^{\mathbf{-}\mathbf{5}\mathbf{r}}\mathbf{+}{\mathit{c}}_{\mathbf{2}}{\mathit{e}}^{\mathbf{r}}\mathbf{+}{\mathit{c}}_{\mathbf{3}}\mathit{r}{\mathit{e}}^{\mathbf{r}}\mathbf{+}{\mathit{c}}_{\mathbf{4}}{\mathit{e}}^{\mathbf{-}\mathbf{r}}\mathbf{+}{\mathit{c}}_{\mathbf{5}}\mathit{r}{\mathit{e}}^{\mathbf{-}\mathbf{r}}$