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Term Test 2 / Re: Problem 2 (noon)
« on: November 19, 2019, 04:53:54 AM »
(a) By Abel's identity, W($y_{1}, y_{2}, y_{3}$)(t) = cexp($\int -1dt$) = ce$^{-t}$
(b) The characteristic equation reads $$ r^{3} + r^{2} + 4r + 4 = (r+1)(r^{2} + 4)$$
Which has solutions r = -1, 2i, -2i. Hence, the solutions to the homogenous equation are:$$y_{1} = e^{-t},\; y_{2} = cos(2t), \;y_{3} = sin(2t)$$
so $$W = det(\begin{bmatrix} e^{-t} & cos(2t) & sin(2t) \\ -e^{-t} & -2sin(2t) & 2cos(2t) \\ e^{-t} & -4cos(2t) & -4sin(2t) \end{bmatrix}) = 10e^{-t}$$
So c in part a is 10.
(c) The form of the particular solution is A$e^{-2t}$, so
$$-8Ae^{-2t} +4Ae^{-2t} -8Ae^{-2t} + 4Ae^{-2t} = -24e^{-2t} \Rightarrow -8Ae^{-2t} = -24e^{-2t} \Rightarrow A = 3$$
hence, the solution is $$y = c_{1}e^{-t} + c_{2}cos(2t) + c_{3}sin(2t) + 3e^{-2t}$$
OK, except LaTeX sucks:
2) "operators" should be escaped: \cos, \sin, \tan, \ln
$$
\boxed{y= 3e^{-2t} + C_1e^{-t} +C_2\cos(2t) +C_3\sin(2t).}
$$
(b) The characteristic equation reads $$ r^{3} + r^{2} + 4r + 4 = (r+1)(r^{2} + 4)$$
Which has solutions r = -1, 2i, -2i. Hence, the solutions to the homogenous equation are:$$y_{1} = e^{-t},\; y_{2} = cos(2t), \;y_{3} = sin(2t)$$
so $$W = det(\begin{bmatrix} e^{-t} & cos(2t) & sin(2t) \\ -e^{-t} & -2sin(2t) & 2cos(2t) \\ e^{-t} & -4cos(2t) & -4sin(2t) \end{bmatrix}) = 10e^{-t}$$
So c in part a is 10.
(c) The form of the particular solution is A$e^{-2t}$, so
$$-8Ae^{-2t} +4Ae^{-2t} -8Ae^{-2t} + 4Ae^{-2t} = -24e^{-2t} \Rightarrow -8Ae^{-2t} = -24e^{-2t} \Rightarrow A = 3$$
hence, the solution is $$y = c_{1}e^{-t} + c_{2}cos(2t) + c_{3}sin(2t) + 3e^{-2t}$$
OK, except LaTeX sucks:
2) "operators" should be escaped: \cos, \sin, \tan, \ln
$$
\boxed{y= 3e^{-2t} + C_1e^{-t} +C_2\cos(2t) +C_3\sin(2t).}
$$