Question

Solve the given system of differential equations by systematic elimination.

Dx	+	D2y	=	e3t
(D + 1)x	+	(D − 1)y	=	4e3t

Answer 1

Solution:

$\dpi{100} \small Dx+D^{2}y=e^{3t}................\left ( i \right )$

$\dpi{100} \small \left ( D+1 \right )x+\left ( D-1 \right )y=4e^{3t}...............\left ( ii \right )$

Multiply equation $\dpi{100} \small \left ( i \right )$ by $\dpi{100} \small \left ( D+1 \right )$    and equation $\dpi{100} \small \left ( ii \right )$    by $\dpi{100} \small D$    and subtract the resulting equations

$\dpi{100} \small D\left ( D+1 \right )x+D^{2}\left ( D+1 \right )y=\left ( D+1 \right )e^{3t}$

$\dpi{100} \small D\left ( D+1 \right )x+D\left ( D-1 \right )y=4De^{3t}$

$\dpi{100} \small -$                  $\dpi{100} \small -$                         $\dpi{100} \small -$

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$\dpi{100} \small 0x+\left ( D^{3}+D \right )y=\left ( D+1 \right )e^{3t}-4De^{3t}$

$\dpi{100} \small \therefore \left ( D^{3}+D \right )y=3e^{3t}+e^{3t}-12e^{3t}=-8e^{3t}$

The Auxiliary equation is $\dpi{100} \small D^{3}+D=0$

$\dpi{100} \small D\left ( D^{2} +1\right )=0\Rightarrow D=0,\pm i$

The complementary function is   $\dpi{100} \small y_{c}=c_{1}+c_{2}cost+c_{3}sint$

Particular integral $\dpi{100} \small \left ( y_{p} \right )$

$\dpi{100} \small =\frac{1}{f\left ( D \right )}\left ( -8e^{3t} \right )$

$\dpi{100} \small =\frac{-8}{D^{3}+D}\left ( e^{3t} \right )$

$\dpi{100} \small f\left ( D \right )=D^{3}+D$

$\dpi{100} \small \therefore f\left ( 3 \right )=3^{3}+3=27+3=30\neq 0$

$\dpi{100} \small \therefore y_{p}=\frac{-8}{30}e^{3t}=\frac{-4}{15}e^{3t}$

$\dpi{100} \small \therefore y=y_{c}+y_{p}$

$\dpi{100} \small \therefore y=c_{1}+c_{2}cost+c_{3}sint-\frac{4}{15}e^{3t}$

Substitute    $\dpi{100} \small y=c_{1}+c_{2}cost+c_{3}sint-\frac{4}{15}e^{3t}$ in equation $\dpi{100} \small \left ( i \right )$

$\dpi{100} \small Dx+D^{2}\left ( c_{1}+c_{2}cost+c_{3}sint-\frac{4}{15}e^{3t}\right )=e^{3t}$

$\dpi{100} \small \therefore Dx-c_{2}cost-c_{3}sint-\frac{12}{5}e^{3t}=e^{3t}$

$\dpi{100} \small \therefore Dx=c_{2}cost+c_{3}sint+\frac{17}{5}e^{3t}$

$\dpi{100} \small \therefore x=\int \left ( c_{2}cost+c_{3}sint+\frac{17}{5}e^{3t} \right )dt$

$\dpi{100} \small \therefore x= c_{2}sint-c_{3}cost+\frac{17}{15}e^{3t}$

$\dpi{100} \small \therefore$The solution to the system are

$\dpi{100} \small x= c_{2}sint-c_{3}cost+\frac{17}{15}e^{3t}\: ,\: \: \: \small y=c_{1}+c_{2}cost+c_{3}sint-\frac{4}{15}e^{3t}$