Following are the solution to the system equation:
[tex](D+5)x+D^2y = \cos\ t--- (1)\\\\ D^2x+(D+9)\ y=t^2 + 3t ---(2)\\\\[/tex]
Multiply the Equation (1) by D+9 and Equation (2) by D^2\\\\
Then
[tex](D+9) (D+5)x+(D+9) D^2y = (D+9) \cos\ t \\\\D^4 \ x+ D^2 (D+9)\ y=D^2 (t^2 + 3t)\\\\\[(D+9) (D+5)-D^4]x=(D+9) \cos\ t-D^2 (t^2+3t)\\\\\to [D^2 +5D+9D+45-D^4]x=-\sin t +9 \cos t-2\\\\\to [D^2 +13D+45-D^4] x = -\sin t +9 \cos t-2\\\\ \to (D^4 -D^2-13D-45) x = \sin t -9 \cos t+2\\\\\to x(t) = \frac{\sin\ t -9 \cos t+2}{D^4-D^2-13 D-45}\\\\[/tex]
[tex]x(t) = \frac{\sin\ t }{D^4-D^2-13 D-45} - 9 \frac{\cos t}{D^4-D^2-13 D-45} + \frac{2}{D^4-D^2-13 D-45}\\\\[/tex]
[tex]=- \frac{\sin\ t}{13D+43} +9 \frac{\cos \ t}{13D +43} -\frac{2}{45} \\\\=-\frac{ (13D-43)\sin t}{(13D-43)(13D+43)} +\frac{9(13D-43) \cos t}{169D^2-1949 } -\frac{2}{45}\\\\[/tex]
by solving the value we get
[tex]=\frac{1}{2018} (13 \cos t -43 \sin t) -\frac{9}{2018} (-13\sin t- 43 \cos t) - \frac{2}{45}\\\\[/tex]
Learn more information about the equation here:
brainly.com/question/3336779
