Question

In Problems69 and 70 use a CAS as an aid in solving the auxiliary equation. Form the general solution of the differential equation. Then use a CAS as an aid in solving the system of equations for the coefficients ci, i = 1, 2, 3, 4 that results when the initial conditions are applied to the general solution

$$\begin{gathered} \mathbf{2}{\mathit{y}}^{\left(4\right)}\mathbf{+}\mathbf{3}\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{16}\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{15}\mathit{y}\mathbf{\text{'}}\mathbf{-}\mathbf{4}\mathit{y}\mathbf{=}\mathbf{0} \\ \mathit{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{-}\mathbf{2}\mathbf{,}\mathit{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{6}\mathbf{,}\mathbf{}\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{3}\mathbf{,}\mathbf{}\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}} \end{gathered}$$

Short answer

$y=\frac{918}{25}{e}^x-\frac{4}{75}{e}^{-4x}-\frac{116}{3}{e}^{\frac{1}{2}x}-\frac{58}{5}x{e}^x$

Step-by-step solution

Finding the auxiliary equation of the given differential equation

We have a third-order differential equation on our hands.

$2y^{\left(4\right)}+3y\text{'}-16y\text{'}\text{'}+15y\text{'}-4y=0......\left(1\right)$

The given initial conditions are

$y\left(0\right)=-2,y\text{'}\left(0\right)=6,y\text{'}\text{'}\left(0\right)=3,y\text{'}\text{'}\text{'}\left(0\right)=\frac{1}{2}$

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

Substitute$y=e^{mx},y\text{'}=m{e}^{mx}$ and$y\text{'}\text{'}=m^2{e}^{mx},y\text{'}\text{'}\text{'}=m^3{e}^{mx}\mathrm{and}{y}^{\left(4\right)}=m^4{e}^{mx}$ into role="math" localid="1663919790075" $2y^{\left(4\right)}+3y\text{'}-16y\text{'}\text{'}+15y\text{'}-4y=0$ to obtain the auxiliary equation.

$\begin{array}{c}2m^4{e}^{mx}+3m^3{e}^{mx}-16m^2{e}^{mx}+15m{e}^{mx}-4e^{mx}=0\\ e^{mx}(2m^4+3m^3-16m^2+15m-4)=0\end{array}$

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

$2m^4+3m^3-16m^2+15m-4=0$is the auxiliary equation for the given differential

Equation

Obtaining the arbitrary constants in the given differential equation

This auxiliary equation is difficult to solve; nevertheless, we can solve it using a

computer algebra system and extract the roots as follows:

$m_1=1,m_2=1,m_3=-4,m_4=\frac{1}{2}$ which are real and repeated.

We must acquire an additional linearly independent solution for each repeated root

because we have repeated roots, and then we can derive the general solution of this

homogeneous equation $2y^4+3y\text{'}-16y\text{'}\text{'}+15y\text{'}-4y=0$ as

$\begin{array}{c}y=n{e}^{m_{1}x}(c+sx)+k_3{e}^{m_{3}x}+k_4{e}^{m_{4}x}\\ y=nc{e}^x+nsx{e}^x+k_3{e}^{-4x}+k_4{e}^{\frac{1}{2}x}\\ y=k_1{e}^x+k_{2}x{e}^x+k_3{e}^{-4x}+k_4{e}^{\frac{1}{2}x}······\left(2\right)\end{array}$

Where$k_1=nc,k_2=ns,k_3\mathrm{and}{k}_4$are arbitrary constants.

Then, using the following technique, we must apply the given beginning conditions of

the given differential equation to obtain the arbitrary constants.

The point (x, y) = (0, -2) must first be applied to the general solution in equation (6) as

follows

$\begin{array}{l}-2=k_1{e}^0+k_2\left(0\right)+k_3{e}^0+k_4{e}^0\\ k_1+k_3+k_4=-2·······\left(\mathrm{a}\right)\end{array}$

Finding the  first derivative

For the general solution of the differential equation (2), we must take the first derivative as follows:

$\begin{array}{c}y\text{'}=k_1{e}^x+k_2{e}^x+k_{2}x{e}^x-4k_3{e}^{-4x}+\frac{1}{2}{k}_4{e}^{\frac{1}{2}x}\\ y\text{'}=k_1{e}^x+k_2(x+1)e^x-4k_3{e}^{-4x}+\frac{1}{2}{k}_4{e}^{\frac{1}{2}x}······\left(3\right)\end{array}$

Then, in equation (3), apply the point to the first derivative as

$\begin{array}{l}6=k_1{e}^0+k_2(0+1)e^0-4k_3{e}^0+\frac{1}{2}{k}_4{e}^0\\ k_1+k_2-4k_3+\frac{1}{2}{k}_4-6\\ 2k_1+2k_2-8k_3+k_4=12······\left(\mathrm{b}\right)\end{array}$

Finding the second derivative

For the general solution of the differential equation (2), we must take the second derivative as follows:

$\begin{array}{c}y\text{'}\text{'}=k_1{e}^x+k_2{e}^x+k_2(x+1)e^x+16k_3{e}^{-4x}+\frac{1}{4}{k}_4{e}^{\frac{1}{2}x}\\ y\text{'}\text{'}=k_1{e}^x+k_2(x+2)e^x+16k_3{e}^{-4x}+\frac{1}{4}{k}_4{e}^{\frac{1}{2}x}······\left(4\right)\end{array}$

Then, in equation (4), apply the point $(x,y\text{'}\text{'})=(0,3)$ to the first derivative as follows:

$\begin{array}{l}3=k_1{e}^0+k_2(0+2)e^0+16k_3{e}^0+\frac{1}{4}{k}_4{e}^0\\ k_1+2k_2+16k_3+\frac{1}{4}{k}_4=3\\ 4k_1+8k_2+64k_3+k_4=12······\left(\mathrm{c}\right)\end{array}$

Finding the third derivative

For the general solution of the differential equation (2), we must take the third derivative as follows:

$\begin{array}{c}y\text{'}\text{'}\text{'}=k_1{e}^x+k_2{e}^x+k_2(x+2)e^x-64k_3{e}^{-4x}+\frac{1}{8}{k}_4{e}^{\frac{1}{2}x}\\ y\text{'}\text{'}\text{'}=k_1{e}^x+k_2(x+3)e^x-64k_3{e}^{-4x}+\frac{1}{8}{k}_4{e}^{\frac{1}{2}x}·····\left(5\right)\end{array}$

Then, in equation (5), apply the point $(x,y\text{'}\text{'}\text{'})=\left(0,\frac{1}{2}\right)$ to the first derivative as follows:

$\begin{array}{l}\frac{1}{2}=k_1{e}^0+k_2(0+3)e^x-64k_3{e}^0+\frac{1}{8}{k}_4{e}^0\\ k_1+3k_2-64k_3+\frac{1}{8}{k}_4=\frac{1}{2}\\ 8k_1+24k_2-512k_3+k_4=4······\left(\mathrm{d}\right)\end{array}$

The value of the arbitrary constants can be obtained by solving equations (a), (b), (c), and (d) using CAS.

$k_1=\frac{918}{25},k_2=-\frac{58}{5},k_3=-\frac{4}{75},k_4=-\frac{116}{3}$.

Finally, in the general solution shown in equation (2), insert the values of arbitrary

constants, and we get $y=\frac{918}{25}{e}^x-\frac{4}{75}{e}^{-4x}-\frac{116}{3}{e}^{\frac{1}{2}x}-\frac{58}{5}x{e}^x$

Hence, the solution is $y=\frac{918}{25}{e}^x-\frac{4}{75}{e}^{-4x}-\frac{116}{3}{e}^{\frac{1}{2}x}-\frac{58}{5}x{e}^x$