Question

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

${\mathit{y}}^{\mathbf{\left(}\mathbf{4}\mathbf{\right)}}\mathbf{-}\mathbf{2}\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathit{y}\mathbf{=}\mathbf{0}$

Short answer

$y=c_1{e}^x+c_{2}x{e}^x+c_3{e}^{-x}+c_{4}x{e}^{-x}$

Step-by-step solution

Determine the general equation of the higher – order differential equation

We have the fourth order differential equation

$y^{\left(4\right)}-2y^{\text{'}\text{'}}+y=0$

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.

$$\begin{gathered} m^4{e}^{mx}-2m^2{e}^{mx}+e^{mx}=0 \\ {e}^{mx}\left(m^4-2m^2+1\right)=0 \end{gathered}$$

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

$$\begin{gathered} m^4-2m^2+1=0 \\ \left(m^2-1\right)\left(m^2-1\right)=0 \\ (m-1)(m+1)(m-1)(m+1)=0 \\ (m-1{)}^2(m+1{)}^2=0 \end{gathered}$$

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

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

$y=n_1{e}^x\left(k_1+s_{1}x\right)+n_2{e}^{-x}\left(k_2+s_{2}x\right)=c_1{e}^x+c_{2}x{e}^x+c_3{e}^{-x}+c_{4}x{e}^{-x}$

Where we have $c_1=n_1{k}_1,c_2=n_1{s}_1,c_3=n_2{k}_2\text{and}{c}_4=n_2{s}_2$are arbitrary constants.

The general solution is$\mathit{y}\mathbf{=}{\mathit{c}}_{\mathbf{1}}{\mathit{e}}^{\mathbf{x}}\mathbf{+}{\mathit{c}}_{\mathbf{2}}\mathit{x}{\mathit{e}}^{\mathbf{x}}\mathbf{+}{\mathit{c}}_{\mathbf{3}}{\mathit{e}}^{\mathbf{-}\mathbf{x}}\mathbf{+}{\mathit{c}}_{\mathbf{4}}\mathit{x}{\mathit{e}}^{\mathbf{-}\mathbf{x}}$