Toronto Math Forum
MAT2442019F => MAT244Test & Quizzes => Term Test 2 => Topic started by: Victor Ivrii on November 19, 2019, 04:19:53 AM

Consider equation
\begin{equation}
y'''+y''+4y'+4y=24e^{2t}.
\label{21}
\end{equation}
(a) Write a differential equation for Wronskian of $y_1,y_2,y_3$, which are solutions for homogeneous equation and solve it.
(b) Find fundamental system $\{y_1,y_2,y_3\}$ of solutions for homogeneous equation, and find their Wronskian. Compare with (a).
(c) Find the general solution of (\ref{21}).

(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
$$
\boxed{y= 3e^{2t} + C_1e^{t} +C_2\cos(2t) +C_3\sin(2t).}
$$

a)W=ce^{∫p(t)dt} with p(t)=1.
Thus, W=ce^{t}

Please see the attachment for the answer

answer for tt2 question2

Here is my solution :)