MAT244--2018F > Final Exam


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Qinger Zhang:

--- Quote from: Xu Zihan on December 14, 2018, 10:34:57 AM ---
--- Quote from: Qi Cui on December 14, 2018, 08:52:58 AM ---$$Homo: r^3 -3r^2+4r-2=0$$
$$\therefore y_c(t)=c_1e^t+c_2e^tcost+c_3e^tsint$$
$$y^{'''}-3y{''}+4y{'}-2y=10e^t+10e^{-t} +20cost$$
$$y^{'}=Ate^t+ Ae^t $$
$$y^{''} = Ate^t+ 2Ae^t $$
$$y^{'}= Ate^t+ 3Ae^t  $$
substitute above into the function:
$$we have: A=-10$$
$$\therefore y_{p1}(t)=-10e^t $$
$$y_{p2}(t)=Ae^{-t} $$
$$y^{'}= -Ae^{-t}$$
$$y^{''}= Ae^{-t} $$
$$y^{'''}= -Ae^{-t} $$
substitute above into the function:
 I think A should be -1 here.
$$\therefore y_{p2}(t)=-e^{-t}  $$
$$y_{p3}(t)=Acost+Bsint $$
$$y^{'}= -Asint+Bcost$$
$$y^{''}= -Acost-Bsint $$
$$y^{'''}= Asint-Bcost$$
substitute above into the function:
we have A=2, B=6
$$\therefore y_{p3}(t)=2cost+6sint $$
$$\therefore y(t)= c_1e^t+c_2e^tcost+c_3e^tsint-10e^t- e^{-t}  + 2cost+6sint $$

--- End quote ---
$$ y_{p2}(t)=-e^{-t}  $$ because W2 is negative

--- End quote ---

Victor Ivrii:
Writing characteristic equation: $L(k):= k^3-3k^2+4k-2=0$. Obviously, one root $k_1=1$; then $k_2+k_3= 2$, $k_1k_2=2$ and they satisfy $k^2-2k+2=0\implies k_{1,2}= 1\pm i$. Then
y^*= C_1e^t + C_2 e^t \cos(t) + C_3 e^t\sin(t)
is a general solution to the homogeneous equation.

Solving inhomogeneous equations with RHE $f_1=10e^{t}$, $f_2=10 e^{-t}$, $f_3=\cos(t)$:
&y_{p1}= At e^t,\\
&y_{p3}= C\cos(t)+D\sin(t).
Here $A L'(k)|_{k=1} = A(3k^2-6k+4)|_{k=1}=10\implies A=10$,
$BL(-1) =-10 B=10\implies B=-1$ and
(C+iD)L(i)= (A+iB) (1+3i)=20\implies
C+iD= \frac{20}{1+3i}=\frac{20(1-3i)}{10}=
2-6i\implies C=2, D=6.
&y_{p1}= 10t e^t,\\
&y_{p3}= 2\cos(t)+6\sin(t).
y= \underbracket{10t e^{t}}_{y_{p1}}\underbracket{-e^{-t}}_{y_{p2}}+ \underbracket{2\cos(t)+6\sin(t)}_{y_{p3}} + \underbracket{C_1e^t + C_2 e^t \cos(t) + C_3 e^t\sin(t)}_{y^*}.


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