Question

In Problems 1–20 solve the given system of differential equations bysystematic elimination.

$$\begin{gathered} \frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{x}}{\mathbf{d}{\mathbf{t}}^{\mathbf{2}}}\mathbf{=}\mathbf{4}\mathit{y}\mathbf{+}{\mathit{e}}^{\mathbf{t}} \\ \frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{y}}{\mathbf{d}{\mathbf{t}}^{\mathbf{2}}}\mathbf{=}\mathbf{4}\mathit{x}\mathbf{-}{\mathit{e}}^{\mathbf{t}} \end{gathered}$$

Short answer

The solution is,

$x\left(t\right)=c_1{e}^{2t}+c_2{e}^{-2t}+\frac{1}{5}{e}^t+c_3\cos 2t+c_4\sin 2t$and

$y\left(t\right)=c_1{e}^{2t}+c_2{e}^{-2t}-\frac{1}{5}{e}^t-c_3\cos 2t-c_4\sin 2t$

Step-by-step solution

Step 1:Definition of differential equation.

An equation containing the derivatives of one or more unknown functions (or dependent variables), with respect to one or more independent variables, is said to be a differential equation (DE).

Differential operator notation.

We have the system of first order equations

$$\begin{gathered} \frac{d^{2}x}{d{t}^2}=4y+e^t \\ \frac{d^{2}y}{d{t}^2}=4x-e^t \end{gathered}$$

Which is

$$\begin{gathered} x\text{'}\text{'}-4y=e^t \\ y\text{'}\text{'}-4x=e^t \end{gathered}$$

Which by using differential operating is equivalent to

$$\begin{gathered} D^{2}x-4y=e^t...............\left(1\right) \\ 4x-D^{2}y=e^t...............\left(2\right) \end{gathered}$$

We have to solve it using systematic elimination as the following technique:

Operate the first equation by the operator $D^2$and multiply the second equation by -4, then we obtain

$$\begin{gathered} D^{4}x-4D^{2}y=D^2{e}^t...............\left(1a\right) \\ -16x+4D^{2}y=-4e^t...............\left(2a\right) \end{gathered}$$

After that, add equation (1a) to equation (2a) and simplify, then we obtain

$$\begin{gathered} D^{4}x-4D^{2}y-16x+4D^{2}y=D^2{e}^t-4e^t \\ (D^4-16)x=e^t-4e^t \\ (D^4-16)x=-3e^t...................\left(3\right) \\ (D^2-4)(D^2+4)x=-3e^t \end{gathered}$$

After that we have to obtain the solution of the corresponding homogeneous equation $(D^2-4)(D^2+4)x=0$as the following:

We have the auxiliary solution for x as,

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

Then we have the roots,

$m_1=2$and $m_2=-2$and $m_{3,4}=\pm 2i$

Then we can obtain the homogenous solution for x as,

$x_c\left(t\right)=c_1{e}^{2t}+c_2{e}^{-2t}+c_3\cos 2t+c_4\sin 2t.........\left(4\right)$

To find the particular solutions.

We have to obtain its particular solution by assuming that $x_p=A{e}^t$, then differentiate with respect to t as,

$$\begin{gathered} D{x}_p=A{e}^t \\ {D}^2{x}_p=A{e}^t \\ {D}^3{x}_p=A{e}^t \\ {D}^4{x}_p=A{e}^t.................\left(5\right) \end{gathered}$$

Then substitute with $x_p=A{e}^t$and the fourth derivative shown in equation (5) into equation (3), then we get

$$\begin{gathered} A{e}^t-16A{e}^t=-3e^t \\ -15A{e}^t=-3e^t \end{gathered}$$

Simplify,

$$\begin{gathered} A=\frac{-3}{-15} \\ A=\frac{1}{5} \end{gathered}$$

Now we obtain the particular solution for x as,

$x_p=\frac{1}{5}{e}^t.................\left(6\right)$

Then from equation(4) and (6) we can obtain the general solution for x as,

$x\left(t\right)=x_c+x_p$

$x\left(t\right)=c_1{e}^{2t}+c_2{e}^{-2t}+c_3\cos 2t+c_4\sin 2t+\frac{1}{5}{e}^t$

$x\left(t\right)=c_1{e}^{2t}+c_2{e}^{-2t}+\frac{1}{5}{e}^t+c_3\cos 2t+c_4\sin 2t$...............(7)

To find the general solutions.

Now we have to obtain the general solution for y as the following:

Differentiate the solution of x shown in equation (7) as

$$\begin{gathered} Dx=2c_1{e}^{2t}-2c_2{e}^{-2t}+\frac{1}{5}{e}^t-2c_3\sin 2t+2c_4\cos 2t \\ {D}^{2}x=4c_1{e}^{2t}+4c_2{e}^{-2t}+\frac{1}{5}{e}^t-4c_3\cos 2t-4c_4\sin 2t.........\left(8\right) \end{gathered}$$

After that, substitute with equation (8) into equation (1), then we obtain

$$\begin{gathered} 4c_1{e}^{2t}+4c_2{e}^{-2t}+\frac{1}{5}{e}^t-4c_3\cos 2t-4c_4\sin 2t-4y=e^t \\ 4y=4c_1{e}^{2t}+4c_2{e}^{-2t}+\frac{1}{5}{e}^t-4c_3\cos 2t-4c_4\sin 2t-e^t \\ 4y=4c_1{e}^{2t}+4c_2{e}^{-2t}-\frac{4}{5}{e}^t-4c_3\cos 2t-4c_4\sin 2t \end{gathered}$$

Then we obtain,

$y\left(t\right)=\frac{1}{4}(4c_1{e}^{2t}+4c_2{e}^{-2t}-\frac{4}{5}{e}^t-4c_3\cos 2t-4c_4\sin 2t)$

$y\left(t\right)=c_1{e}^{2t}+c_2{e}^{-2t}-\frac{1}{5}{e}^t-c_3\cos 2t-c_4\sin 2t$is the general solution for y.

Therefore, a solution of the system is

$x\left(t\right)=c_1{e}^{2t}+c_2{e}^{-2t}+\frac{1}{5}{e}^t+c_3\cos 2t+c_4\sin 2t$and

$y\left(t\right)=c_1{e}^{2t}+c_2{e}^{-2t}-\frac{1}{5}{e}^t-c_3\cos 2t-c_4\sin 2t$