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

$\begin{array}{l}{\mathbf{y}}^{\left(4\right)}\mathbf{-}\mathbf{3}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{3}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{y}\mathbf{\text{'}}\mathbf{=}\mathbf{0}\\ \mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{ }\mathbf{,}\mathbf{y}\mathbf{\text{'}}\left(0\right)\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{ }\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}\left(0\right)\mathbf{=}\mathbf{1}\end{array}$

Short answer

$y=2-2e^x+2x{e}^x-\frac{1}{2}{x}^2{e}^x$

Step-by-step solution

Definition of general equation

We previously concluded that the given system's is general solution then its used the type of matrix form.

The third order differential equation:

$y^{\left(4\right)}-3y\text{'}\text{'}\text{'}+3y\text{'}\text{'}-y\text{'}=0······\left(1\right)$

Initial conditions:

$y\left(0\right)=0 ,y\text{'}\left(0\right)=0, y\text{'}\text{'}\text{'}\left(0\right)=1$

Consider role="math" localid="1663922708015" $\mathit{y}\mathbf{=}{\mathit{e}}^{\mathbf{m}\mathbf{x}}$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 to obtain the auxiliary equation.

$\begin{array}{l}{m}^4 e^{mx}-3m^3{e}^{mx}+3m^2 e^{mx}-m e^{mx}=0\\ e^{mx}\left(m^4-3m^3+3m^3+m\right)=0\end{array}$

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

$\begin{array}{c}{m}^4-3m^3+3m^2+m=0\\ m\left(m^3-3m^2+3m+1\right)=0\end{array}$

Auxiliary equation of the given differential equation.

This auxiliary equation cannot be solved easily, then using Computer Algebra System we can solve it and obtain the roots as

$m_1=0 ,m_2=1 ,m_3=1 \mathrm{and} m_4=1$

which are real and repeated. Since we have repeated roots, then we have to obtain an additional linearly independent solution for each repeated root, then we can obtain the general solution of this homogeneous equation

$y^{\left(4\right)}-3y\text{'}\text{'}\text{'}+3y\text{'}\text{'}-y\text{'}=0$

Arbitrary constants:

$k_1,k_2=n\times a,k_3=n\times b \mathrm{and} k_4=n\times c$

After that, we have to apply the given initial conditions of the given differential equation to obtain the arbitrary constants as the following technique :

First, we have to apply the point $\left(x,y\right)=\left(0,0\right)$into the general solution in equation (2)

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

Second; we have to take the first derivative for the general solution equation (2) as

$\begin{array}{c}y\text{'}=k_2{e}^x+k_3 e^x+k_{3 }x{e}^x+k_4 x^2{e}^x+2k_4 x{e}^x\\ =k_2{e}^x+k_3\left(x+1\right)e^x+k_4\left(x^2+2x\right)e^x······\left(3\right)\end{array}$

Apply the point $\left(x,y\right)=\left(0,0\right)$into first derivative equation (3) as

$\begin{array}{c}0=k_2{e}^0+k_3\left(0+1\right)e^0+k_4\left(0\right)\\ k_2+k_3=0······\left(\mathrm{b}\right)\end{array}$

Find the general solution of the homogeneous equation

Homogeneous Linear Second Order Equation is a linear combination of $y_1\mathrm{and}{y}_2$are real and distinct.

Third, we have to take the second derivative for the general solution of the differential equation(6) as

$\begin{array}{c}y\text{'}\text{'}=k_2{e}^x+k_3 e^x+k_{3 }\left(x+1\right)e^x+k_4\left(x^2+2x\right)e^x+k_4\left(2x+2\right)e^x\\ =k_2{e}^x+k_3 \left(x+2\right)e^x+k_4\left(x^2+4x+2\right)e^x······\left(4\right)\end{array}$

Apply the point:$\left(x,y\text{'}\text{'}\right)=\left(0,1\right)$ into the first derivative in equation (4) as

$\begin{array}{l}1=k_2{e}^0+k_3\left(0+2\right)e^0+k_4\left(0+2\right)e^0\\ k_2+2k_3+ 2k_4=1······\left(\mathrm{c}\right)\end{array}$

Fourth, we have to take the third derivative for the general solution of the differential equation as

$\begin{array}{c}y\text{'}\text{'}\text{'}=k_2{e}^x+k_3 e^x+k_{3 }\left(x+2\right)e^x+k_4\left(x^2+4x+2\right)e^x+k_4\left(2x+4\right)e^x\\ =k_2{e}^x+k_3 \left(x+3\right)e^x+k_4\left(x^2+6x+6\right)e^x······\left(5\right)\end{array}$

Apply the point:$\left(x,y\text{'}\text{'}\right)=\left(0,1\right)$ into the first derivative in equation (5) as

$\begin{array}{l}1=k_2{e}^0+k_3\left(0+3\right)e^0+k_4\left(0+6\right)e^x\\ k_2+3k_3+ 6k_4=1······\left(\mathrm{d}\right)\end{array}$

Then by solving equations $\left(a\right),\left(b\right),\left(c\right) and\left(d\right)$ we can obtain the value of the arbitrary constants as

$k_1=2 , k_2=-2 ,k_3=2\mathrm{and} k_4=-\frac{1}{2}$

Finally, substitute with the value of arbitrary constants into the general solution showed in equation (2), then we can have

$y=2-2e^x+2x{e}^x-\frac{1}{2}{x}^2{e}^x$

Hence the solution is$y=2-2e^x+2x{e}^x-\frac{1}{2}{x}^2{e}^x$