Question

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

$$\begin{gathered} \frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{x}}{\mathbf{d}{\mathbf{t}}^{\mathbf{2}}}\mathbf{+}\frac{\mathbf{d}\mathbf{y}}{\mathbf{d}\mathbf{t}}\mathbf{=}\mathbf{-}\mathbf{5}\mathit{x} \\ \frac{\mathbf{d}\mathbf{x}}{\mathbf{d}\mathbf{t}}\mathbf{+}\frac{\mathbf{d}\mathbf{y}}{\mathbf{d}\mathbf{t}}\mathbf{=}\mathbf{-}\mathit{x}\mathbf{+}\mathbf{4}\mathit{y} \end{gathered}$$

Short answer

The solution is,

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

$y\left(t\right)=-6c_1{e}^{5t}+\frac{1}{2}{c}_3\cos 2t-\frac{1}{2}{c}_2\sin 2t$

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).

using elimination.

Rewrite the system equations using differential operator

$$\begin{gathered} (D^2+5)x+Dy=0............\left(1\right) \\ (D+1)x+(D-4)y=0..............\left(2\right) \end{gathered}$$

Using elimination to get ,multiply equation (1) by$(D-4)$ and equation (2) by$D$ .

$$\begin{gathered} (D-4)(D^2+5)x+D(D-4)y=0............\left(3\right) \\ D(D+1)x+D(D-4)y=0..............\left(4\right) \end{gathered}$$

Subtract (3) and (4),

$$\begin{gathered} (D^3+5D-4D^2-20-D^2-D)x=0 \\ (D-5)(D^2+4)x=0.......................\left(5\right) \end{gathered}$$

From the auxiliary equation of (5), the roots are,

$m=5,\pm 2i$

Then we can obtain the homogenous solution for x as,

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

using elimination.

Using elimination to get $y\left(t\right)$,multiply equation (1) by $(D+1)$and equation (2) by$(D^2+5)$ .

$$\begin{gathered} (D+1)(D^2+5)x+D(D+1)y=0............\left(6\right) \\ (D^2+5)(D+1)x+(D^2+5)(D-4)y=0..............\left(7\right) \end{gathered}$$

Subtract (6) and (7),

$$\begin{gathered} (D^2+D-D^3+4D^2-5D+20)y=0 \\ -(D-5)(D^2+4)y=0.......................\left(8\right) \end{gathered}$$

From the auxiliary equation of (8), the roots are,

$m=5,\pm 2i$

Then we can obtain the homogenous solution for y as,

$y\left(t\right)=c_4{e}^{5t}+c_5\cos 2t+c_6\sin 2t$

To find the general solutions:

Substitute with $x\left(t\right)$and $y\left(t\right)$into (2),

$(6c+c_4{)}_1{e}^{5t}+(c_2+2c_3-4c_5+2c_6)\cos 2t+(-2c_2+c_3-2c_5-4c_6)\sin 2t=0$

Equate the constants we get,

$$\begin{gathered} c_4=-6c_1 \\ {c}_5=\frac{1}{2}{c}_3 \\ {c}_6=-\frac{1}{2}{c}_2 \end{gathered}$$

Hence,

$y\left(t\right)=-6c_1{e}^{5t}+\frac{1}{2}{c}_3\cos 2t-\frac{1}{2}{c}_2\sin 2t$

Therefore, a solution of the system is

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

$y\left(t\right)=-6c_1{e}^{5t}+\frac{1}{2}{c}_3\cos 2t-\frac{1}{2}{c}_2\sin 2t$