In Problems 3 – 18, use the elimination method to find a general solution for the given linear system, where differentiation is with respect to t.

$$\begin{gathered} \frac{\mathbf{dw}}{\mathbf{dt}}\mathbf{=}\mathbf{5}\mathbf{w}\mathbf{+}\mathbf{2}\mathbf{z}\mathbf{+}\mathbf{5}\mathbf{t}\mathbf{,} \\ \frac{\mathbf{dz}}{\mathbf{dt}}\mathbf{=}\mathbf{3}\mathbf{w}\mathbf{+}\mathbf{4}\mathbf{z}\mathbf{+}\mathbf{17}\mathbf{t} \end{gathered}$$

Elimination Procedure for 2 × 2 Systems

To find a general solution for the system

$$\begin{gathered} {\mathbf{L}}_{\mathbf{1}}\left[\mathbf{x}\right]\mathbf{+}{\mathbf{L}}_{\mathbf{2}}\left[\mathbf{y}\right]\mathbf{=}{\mathbf{f}}_{\mathbf{1}}\mathbf{,} \\ {\mathbf{L}}_{\mathbf{3}}\left[\mathbf{x}\right]\mathbf{+}{\mathbf{L}}_{\mathbf{4}}\left[\mathbf{y}\right]\mathbf{=}{\mathbf{f}}_{\mathbf{2}}\mathbf{,} \end{gathered}$$

Where${\mathbf{L}}_{\mathbf{1}}\mathbf{,}{\mathbf{L}}_{\mathbf{2}}\mathbf{,}{\mathbf{L}}_{\mathbf{3}}\mathbf{,}$ and${\mathbf{L}}_{\mathbf{4}}$ are polynomials in$D=\frac{d}{dt}$

Given that,

$$\begin{gathered} \frac{dw}{dt}=5w+2z+5t ......\mathbf{\left(}\mathbf{1}\mathbf{\right)} \\ \frac{dz}{dt}=3w+4z+17t ......\mathbf{\left(}\mathbf{2}\mathbf{\right)} \end{gathered}$$

Let us rewrite the system in operator form,

$$\begin{gathered} \left(D-5\right)\left[w\right]-2z=5t ......\mathbf{\left(}\mathbf{3}\mathbf{\right)} \\ -3w+\left(D-4\right)\left[z\right]=17t ......\mathbf{\left(}\mathbf{4}\mathbf{\right)} \end{gathered}$$

Multiply 3 on equation (3) and D-5 on both sides of equation (4) then add equation (3) and (4) together one gets,

$$\begin{gathered} 3\left(D-5\right)\left[w\right]-6z-3\left(D-5\right)w+\left(D-5\right)\left(D-4\right)\left[z\right]=15t+\left(D-5\right)17t \\ -6z\left(D^2-4D-5D+20\right)\left[z\right]=15t+17-85t \\ \left(D^2-9D+14\right)\left[z\right]=17-70t \\ \left(D^2-9D+14\right)\left[z\right]=17-70t ......\mathbf{\left(}\mathbf{5}\mathbf{\right)} \end{gathered}$$

Since the auxiliary equation to the corresponding homogeneous equation is:$r^2-9r+14=0$

. The roots are r=7 and r=2.

Then, the homogeneous solution of u is:

$z_h\left(t\right)=c_1{e}^{2t}+c_2{e}^{7t} ......\mathbf{\left(}\mathbf{6}\mathbf{\right)}$

Let us take the undetermined coefficients and assume that

$z_p\left(t\right)=A+Bt ......\mathbf{\left(}\mathbf{7}\mathbf{\right)}$

Now derivate the equation (7)

$$\begin{gathered} D\left[z_p\left(t\right)\right]=B \\ {D}^2\left[z_p\left(t\right)\right]=0 \end{gathered}$$

Substitute the derivation in equation (5).

$$\begin{gathered} \left(D^2-9D+14\right)\left[A+Bt\right]=17-70t \\ {D}^2\left[A+Bt\right]-9D\left[A+Bt\right]+14\left[A+Bt\right]=17-70t \\ 0-9B+14A+14Bt=17-70t \\ -9B+14A+14Bt=17-70t \end{gathered}$$

Now, equalize the like terms.

$$\begin{gathered} -9B+14A=17 \\ 14B=-70 \\ B=\frac{-70}{14}=-5 \end{gathered}$$

Then,

$$\begin{gathered} -9\left(-5\right)+14A=17 \\ 45+14A=17 \\ 14A=17-45 \\ A=\frac{-28}{14}=-2 \end{gathered}$$

So,$z_p\left(t\right)=-2-5t ......\mathbf{\left(}\mathbf{8}\mathbf{\right)}$

Use equations (6) and (8) to get,

$$\begin{gathered} z\left(t\right)=z_h\left(t\right)+z_p\left(t\right) \\ z\left(t\right)=c_1{e}^{2t}+c_2{e}^{7t}-2-5t ......\mathbf{\left(}\mathbf{9}\mathbf{\right)} \end{gathered}$$

Now, take equation (4).

$$\begin{gathered} -3w+\left(D-4\right)\left[z\right]=17t \\ -3w=17t-\left(D-4\right)\left[z\right] \\ w=\frac{1}{3}\left(D-4\right)\left[z\right]-\frac{17}{3}t \\ =\frac{1}{3}\left(D-4\right)\left[c_1{e}^{2t}+c_2{e}^{7t}-2-5t\right]-\frac{17}{3}t \end{gathered}$$

$$\begin{gathered} =\frac{1}{3}\left(2c_1{e}^{2t}+7c_2{e}^{7t}-5-4\left(c_1{e}^{2t}+c_2{e}^{7t}-2-5t\right)\right)-\frac{17}{3}t \\ =\frac{2c_1{e}^{2t}}{3}+\frac{7c_2{e}^{7t}}{3}-\frac{5}{3}-\frac{4c_1{e}^{2t}}{3}-\frac{4c_2{e}^{7t}}{3}+\frac{8}{3}+\frac{20t}{3}-\frac{17t}{3} \\ =-\frac{2}{3}{c}_1{e}^{2t}+c_2{e}^{7t}+t+1 \end{gathered}$$

So, the solution is founded.