Question

Use systematic elimination to solve the given system.

$$\begin{gathered} \mathbf{(}\mathbf{D}\mathbf{+}\mathbf{2}\mathbf{)}\mathbf{x}\mathbf{+}\mathbf{(}\mathbf{D}\mathbf{+}\mathbf{1}\mathbf{)}\mathbf{y}\mathbf{=}\mathbf{sin}\mathbf{2}\mathbf{t} \\ \mathbf{5}\mathbf{x}\mathbf{+}\mathbf{(}\mathbf{D}\mathbf{+}\mathbf{3}\mathbf{)}\mathbf{y}\mathbf{=}\mathbf{cos}\mathbf{2}\mathbf{t} \end{gathered}$$

Short answer

So, the required solution is $\mathrm{x}\left(\mathrm{t}\right)=-\frac{1}{5}(3{\mathrm{c}}_2-{\mathrm{c}}_1)\mathrm{sint}-\frac{1}{5}(3{\mathrm{c}}_1+{\mathrm{c}}_2)\mathrm{cost}-\frac{5}{3}\sin 2\mathrm{t}-\frac{1}{3}\cos 2\mathrm{t}$and $\mathrm{y}\left(\mathrm{t}\right)={\mathrm{c}}_1\mathrm{cost}+{\mathrm{c}}_2\mathrm{sint}-\frac{2}{3}\cos 2\mathrm{t}+\frac{7}{3}\sin 2\mathrm{t}$.

Step-by-step solution

Definition of elimination method

In the elimination method, you either add or subtract the equations to get an equation in one variable.

Using elimination method

We have $(D+2)x+(D+1)y=\sin 2t\mathrm{...............}\left(1\right)$

$5x+(D+3)y=\cos 2t\cdots \text{(2)}$

Now, $\left(1\right)\times 5-(D+2)\times \left(2\right)$gives

$5(D+2)x+5(D+1)y=5\sin 2t$

$\text{(-)}\frac{5(D+2)x+(D+2)(D+3)y=2(\cos 2t-\sin 2t)}{5(D+1)y-(D+2)(D+3)y=7\sin 2t-2\cos 2t}$

$\text{or}(D^2+1)y=2\cos 2t-7\sin 2t-\left(3\right)$

Substitution

Thus$y_c=c_1\cos t+c_2\sin t$

Let$y_p=A\cos 2t+B\sin 2t$

Then$y_p^{\text{'}\text{'}}=-4(A\cos 2t+B\sin 2t)$

Substituting$y_p$in (3) gives

$-3A\cos 2t-3B\sin 2t=2\cos 2t-7\sin 2t$

$\Rightarrow -3A=2$

$-3B=-7$

$\therefore A=-\frac{2}{3};B=\frac{7}{3}$

Thus,$y_p=-\frac{2}{3}\cos 2t+\frac{7}{3}\sin 2t$

$y=y_c+y_p$

$=c_1\cos t+c_2\sin t-\frac{2}{3}\cos 2t+\frac{7}{3}\sin 2t$

Solution of the given system

Then$Dy=-c_1\sin t+c_2\cos t+\frac{4}{3}\sin 2t+\frac{14}{3}\cos 2t$

$(D+3)y=-c_1\sin t+c_2\cos t+\frac{4}{3}\sin 2t+\frac{14}{3}\cos 2t$$+3c_1\cos t+3c_2\sin t-2\cos 2t+7\sin 2t$

$=(-c_1+3c_2)\sin t+(3c_1+c_2)\cos t+\frac{25}{3}\sin 2t+\frac{8}{3}\cos 2t$

$\text{From(2),}x=\frac{1}{5}[\cos 2t-(D+3\left)y\right]$

$=-\frac{1}{5}(3c_2-c_1)\sin t-\frac{1}{5}(3c_1+c_2)\cos t-\frac{5}{3}\sin 2t-\frac{1}{3}\cos 2t$

Thus, a solution of the given system is

$x\left(t\right)=-\frac{1}{5}(3c_2-c_1)\sin t-\frac{1}{5}(3c_1+c_2)\cos t-\frac{5}{3}\sin 2t-\frac{1}{3}\cos 2t$

$y\left(t\right)=c_1\cos t+c_2\sin t-\frac{2}{3}\cos 2t+\frac{7}{3}\sin 2t$