Question

Use systematic elimination to solve the given system.

$\mathbf{(}\mathbf{D}\mathbf{-}\mathbf{2}\mathbf{)}\mathbf{x}\mathbf{-}\mathbf{y}\mathbf{=}\mathbf{-}{\mathbf{e}}^t$

$\mathbf{-}\mathbf{3}\mathbf{x}\mathbf{+}\mathbf{(}\mathbf{D}\mathbf{-}\mathbf{4}\mathbf{)}\mathbf{y}\mathbf{=}\mathbf{-}\mathbf{7}{\mathbf{e}}^t$

Short answer

So, the required solution is $x\left(t\right)=c_1{e}^{5t}+c_2{e}^t+t{e}^t$and $y\left(t\right)=3c_1{e}^{5t}-c_2{e}^t-t{e}^t+2e^t$.

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$(D-2)x-y=-e^t\dots -\left(1\right)$

$-3x+(D-4)y=-7e^t\cdots -\left(2\right)$

Now,$(D-4)\times \left(1\right)+\left(2\right)$ gives

$(D-2)(D-4)x-(D-4)y=3e^t$

$(+)\frac{-3x+(D-4)y}{(D-2)(D-4)x-3x}=-7e^t$

$\text{or}(D^2-6D+5)x=-4e^t$

Substitution

Thus,$x_c=c_1{e}^{5t}+c_2{e}^t$

Let $x_p=At{e}^t$

Thenrole="math" $x_p^{\text{'}}=A{e}^t+A_t{e}^t$

$x_p^{\text{'}\text{'}}=2A{e}^t+At{e}^t$

Substituting $X_p$in (3) gives

$2A{e}^t+{\text{Ate}}^t-6{\text{Ae}}^t-6{\text{Ate}}^t+5{\text{Ate}}^t=-4e^t$

$-4A{e}^t=-4e^t$

$4(A-1)e^t=0$

$\text{Since}{e}^t\ne 0,A=1$

$\therefore x_p=t{e}^t$

Solution for the given system

Thus,$x=x_c+x_p$

$=c_1{e}^{5t}+c_2{e}^t+t{e}^t$

Then$Dx=5c_1{e}^{5t}+c_2{e}^t+t{e}^t+e^t$

$(D-2)x=3c_1{e}^{5t}-c_2{e}^t-t{e}^t+e^t$

From (1),$y=(D-2)x+e^t$

$=3c_1{e}^{5t}-c_2{e}^t-t{e}^t+2e^t$

Thus, a solution of the given system is

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

$y\left(t\right)=3c_1{e}^{5t}-c_2{e}^t-t{e}^t+2e^t$