Question

Solve for x in the following set of simultaneous differential equations by using D-operator methods:

(D + 2) x - 3y = 1

-3x + (D + ) y = e^-t

Answer 1

Given the simultaneous differential equations,

$\begin{aligned} (D + 2) x - 3y &= 1\qquad(1)\\ -3x + (D + 2) y &= e^{-t} ~\quad(2) \end{aligned}$

Applying $(D + 2)$ operator for (1), Multiplying (2) by 3 and adding we get,

$\begin{aligned} (D + 2)^2 x - 3(D+2)y &= 2\qquad(1)\\ -9x + 3(D + 2) y &= 3e^{-t} ~~~(2)\\ \hline (D + 2)^2 x -9x &= 2+3e^{-t}\\ (D^2 + 4D + 4 - 9)x & = 2+3e^{-t}\\ (D^2 + 4D -5)x & = 2+3e^{-t}\\ \end{aligned}$

The auxiliary equation is, $m^{2} + 4m - 5 = 0.$ Solving we get $m = -5,1$.

The complementary function for x(t) is, $\text{C.F} = c_1e^{-5t} + c_2 e^{t}$

$\begin{aligned} \text{Particular Integral for }x(t) &= \frac{1}{D^{2}+4D-5} (2+3e^{-t})\\ &=-\frac{2}{5} -\frac{3e^{-t}}{8}\quad(\text{Replacing D by 0 and -1 for first} \\ &\qquad \qquad \qquad \qquad\text{ and second term respectively})\\ \end{aligned}$

$\therefore x(t) = c_{1}e^{-5t}+c_2e^t-\Big(\dfrac{2}{5}+\dfrac{3e^{-t}}{8}\Big)$

From equation (1), $3y = Dx+2x -1$

$\begin{aligned} 3y&=D\Bigg(c_{1}e^{-5t}+c_2e^t-\Big(\dfrac{2}{5}+\dfrac{3e^{-t}}{8}\Big)\Bigg) + 2\Bigg(c_{1}e^{-5t}+c_2e^t-\Big(\dfrac{2}{5}+\dfrac{3e^{-t}}{8}\Big)\Bigg) - 1\\ 3y &= -5c_{1}e^{-5t}+c_2e^t+\dfrac{3e^{-t}}{8} + 2c_{1}e^{-5t}+2c_2e^t-\dfrac{4}{5}+\dfrac{3e^{-t}}{4}- 1\\ 3y &= -3c_{1}e^{-5t}+3c_2e^t-\dfrac{3e^{-t}}{8} -\dfrac{9}{5}\\ \therefore y(t)& = -c_{1}e^{-5t}+c_2e^t-\dfrac{e^{-t}}{8} -\dfrac{3}{5}\\ \end{aligned}$