Question

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

$$\begin{gathered} \mathbf{(}{\mathit{D}}^{\mathbf{2}}\mathbf{+}\mathbf{5}\mathbf{)}\mathit{x}\mathbf{-}\mathbf{2}\mathit{y}\mathbf{=}\mathbf{0} \\ \mathbf{-}\mathbf{2}\mathit{x}\mathbf{+}\mathbf{(}{\mathit{D}}^{\mathbf{2}}\mathbf{+}\mathbf{2}\mathbf{)}\mathit{y}\mathbf{=}\mathbf{0} \end{gathered}$$

Short answer

The solution is,

$x\left(t\right)=c_1\cos t+c_2\sin t+c_3\cos \sqrt{6}t+c_4\sin \sqrt{6}t$

$y\left(t\right)=\frac{1}{2}{c}_1\cos t+\frac{1}{2}{c}_2\sin t-2c_3\cos \sqrt{6}t-2c_4\sin \sqrt{6}t$

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

$$\begin{gathered} (D^2+5)x-2y=0..........\left(1\right) \\ -2x+(D^2+2)y=0............\left(2\right) \end{gathered}$$

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

Operate the first equation by the operator $(D^2+2)$and multiply the equation(2) by 2. Then we obtain

$$\begin{gathered} (D^2+2)(D^2+5)x-2(D^2+2)y=0........\left(3\right) \\ -4x+2(D^2+2)y=0..............\left(4\right) \end{gathered}$$

After that add equation (3) and (4) and simplify then we obtain

$$\begin{gathered} (D^2+2)(D^2+5)x-2(D^2+2)y-4x+2(D^2+2)y=0 \\ (D^2+2)(D^2+5)x-4x=0 \\ (D^4+5D^2+2D^2+10)x-4x=0 \\ (D^4+7D^2+6)x=0 \end{gathered}$$

After that, we have the auxiliary solution for$x$ as

$$\begin{gathered} (m^4+7m^2+6)=0 \\ (m^2+1)(m^2+6)=0 \end{gathered}$$

Which has the roots

$$\begin{gathered} m_{1,2}=\pm i \\ {m}_{3,4}=\pm \sqrt{6}i \end{gathered}$$

Therefore we obtain,

$x\left(t\right)=c_1\cos t+c_2\sin t+c_3\cos \sqrt{6}t+c_4\sin \sqrt{6}t$..........(5)

To find the solutions.

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

We have to expand equation (1) as

$D^{2}x+2x-2y=0$

$x\text{'}\text{'}+2x-2y=0$.................(6)

After that, differentiate the solution of shown in equation (5) with respect to $t$as,

$x\text{'}=-c_1\sin t+c_2\cos t-\sqrt{6}{c}_3\sin \sqrt{6}t+\sqrt{6}{c}_4\cos \sqrt{6}t$

$x\text{'}\text{'}=-c_1\cos t-c_2\sin t-6c_3\cos \sqrt{6}t-6c_4\sin \sqrt{6}t$..............(7)

After that, substitute with equations (5) and (7) into equation (6), then we obtain

$$\begin{gathered} -c_1\cos t-c_2\sin t-6c_3\cos \sqrt{6}t-6c_4\sin \sqrt{6}t+2(c_1\cos t+c_2\sin t+c_3\cos \sqrt{6}t+c_4\sin \sqrt{6}t)-2y=0 \\ 2y=(-c_1+2c_1)\cos t+(-c_2+2c_2)\sin t+(-6c_3+2c_3)+(-6c_4+2c_4)\sin \sqrt{6}t \\ 2y=c_1\cos t+c_2\sin t-4c_3\cos \sqrt{6}t-4c_4\sin \sqrt{6}t \end{gathered}$$

Then we obtain,

$\begin{array}{rcl}y\left(t\right)& =& \frac{1}{2}(c_1\cos t+c_2\sin t-4c_3\cos \sqrt{6}t-4c_4\sin \sqrt{6}t)\\ & =& \frac{1}{2}{c}_1\cos t+\frac{1}{2}{c}_2\sin t-2c_3\cos \sqrt{6}t-2c_4\sin \sqrt{6}t)\end{array}$

$y\left(t\right)=\frac{1}{2}{c}_1\cos t+\frac{1}{2}{c}_2\sin t-2c_3\cos \sqrt{6}t-2c_4\sin \sqrt{6}t$is the general solution for $y$.

Therefore,

$x\left(t\right)=c_1\cos t+c_2\sin t+c_3\cos \sqrt{6}t+c_4\sin \sqrt{6}t$

$y\left(t\right)=\frac{1}{2}{c}_1\cos t+\frac{1}{2}{c}_2\sin t-2c_3\cos \sqrt{6}t-2c_4\sin \sqrt{6}t$