MAT244--2019F > Term Test 2

Problem 2 (main sitting)

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Victor Ivrii:
Consider equation

y'''+4y''+y'-6y=24e^{t}.
\label{2-1}

(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{2-1}).

Yiheng Bian:
(a):
$$W=ce^{-\int{p(t)}}dt\\ W=ce^{-\int{4}dt}\\ W=ce^{-4t}$$

(b):
We can get:
$$r^3+4r^2+r-6=0\\ (r-1)(r+2)(r+3)=0\\ r_1=1,r_2=-2,r_3=-3$$
Therefore:
$$y=c_1e^t+c_2e^{-2t}+c_3e^{-3t}$$
$$\begin{vmatrix} e^t & e^{-2t} & e^{-3t} \\ e^t & -2e^{-2t} & -3e^{-3t} \\ e^t & 4e^{-2t} & 9e^{-3t} \end{vmatrix}=-12e^{-4t}$$
compare with (a) we get
$$c=-12$$

(c):
Let:
$$y_p(t)=Ate^t\\ y'=A(e^t+te^t)\\ y''=A(2e^t+te^t)\\ y'''=A(3e^t+te^t)$$
We take these into equation and get:
$$A(3e^t+te^t)+4A(2e^t+te^t)+A(e^t+te^t)-6Ate^t=24e^t\\ 12Ae^t=24e^t\\ A=2$$
So:
$$y_p(t)=2te^t$$
Therefore:
$$y=c_1e^t+c_2e^{-2t}+c_3e^{-3t}+2te^t$$

OK. V.I.

xuanzhong:
a)Write equation for Wronskian of y1,y2,y3.
$$W=ce^{-\int p(t)dt}=ce^{-\int 4dt}=ce^{-4t}$$
b)Find fundamental system of solutions for honogenuous equation, and find Wronskian. Comapare with (a).
$$r^3+4r^2+r-6=0$$
$$(r-1)(r+2)(r+3)=0$$
$$r=1, r=-2, r=-3$$
$$y_c(t)=c_1e^t+c_2e^{-2t}+c_3e^{-3t}$$
$$W=\begin{vmatrix} e^t & e^{-2t} & e^{-3t}\\ e^t & -2e^{-2t} & -3e^{-3t}\\ e^t & 4e^{-2t} & 9e^{-3t}\\ \end{vmatrix} = 12e^{-4t}$$
similar solution to (a), but $c=12$

(c)Find the general solution.
$$y_p(t)=Ate^t$$
$$y^\prime=Ae^t+Ate^t=Ae^t(1+t)$$
$$y^{\prime\prime}=2Ae^t+Ate^t=Ae^t(2+t)$$
$$y^{\prime\prime\prime}=3Ae^t+Ate^t=Ae^t(3+t)$$
$$y^{\prime\prime\prime}+4y^{\prime\prime}+y^\prime-6y=24e^t$$
$$Ae^t(3+t+8+4t+1+t-6t)=24e^t$$
$$12Ae^t=24e^t$$
$$A=2$$
Hence, $y(t)=c_1e^t+c_2e^{-2t}+c_3e^{-3t}+2te^t$

Yuying Chen:
$\text(a)\\$
$W=ce^{-\int p(t)dt}=ce^{-\int 4dt}=ce^{-4t}\\$
$\text(b)\\$
$r^3+4r^2+r-6=0\\$
$(r-1)(r^2+5r+6)=0\\$
$r=1, r=-2, r=-3\\$
$\therefore y(t)=c_1e^t+c_2e^{-2t}+c_3e^{-3t}\\$
$W=\begin{vmatrix} e^t & e^{-2t} & e^{-3t}\\ e^t & -2e^{-2t} & -3e^{-3t}\\ e^t & 4e^{-2t} & 9e^{-3t}\\ \end{vmatrix}=-12e^{-4t}\\$
$\text{$\therefore c=-12$compare with (a)}\\$
$\text(c)\\$
$y_p(t)=Ate^t\\$
$y_p^{\prime}(t)=Ae^t+Ate^t\\$
$y_p^{\prime\prime}(t)=2Ae^t+Ate^t\\$
$y_p^{\prime\prime\prime}(t)=3Ae^t+Ate^t\\$
$3Ae^t+Ate^t+8Ae^t+4Ate^t+Ae^t+Ate^t-6Ate^t=12Ae^t\\$
$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad 12A=24\\$
$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad A=2$
$\therefore y(t)=c_1e^t+c_2e^{-2t}+c_3e^{-3t}+2te^t$

Xinyu Jing:
(𝑎)
𝑊=$𝑐𝑒^{−∫𝑝(𝑡)𝑑𝑡}$=$𝑐𝑒^{−∫4𝑑𝑡}$=$𝑐𝑒^{−4𝑡}$
(𝑏)
$𝑟^{3}+4𝑟^{2}+𝑟−6$=0
(𝑟−1)($𝑟^{2}$+5𝑟+6)=0
𝑟=1,𝑟=−2,𝑟=−3
∴𝑦(𝑡)=$𝑐_{1}𝑒^{𝑡}+𝑐_{2}𝑒^{−2𝑡}+𝑐_{3}𝑒^{−3𝑡}$

W=\begin{vmatrix}
e^{t} & e^{-2t} &  e^{-3t} \\
e^{t} & -2e^{-2t} & -3e^{-3t} \\
e^{t} & 4e^{-2t} & 9e^{-3t} \\
\end{vmatrix}=−12𝑒−4𝑡
∴𝑐=−12 compare with (a)
(𝑐)
𝑦𝑝(𝑡)=$𝐴𝑡𝑒^{𝑡}$
𝑦′𝑝(𝑡)=$𝐴𝑒^{𝑡}+𝐴𝑡𝑒^{𝑡}$
𝑦′′𝑝(𝑡)=$2𝐴𝑒^{𝑡}+𝐴𝑡𝑒^{𝑡}$
𝑦′′′𝑝(𝑡)=$3𝐴𝑒^{𝑡}+𝐴𝑡𝑒^{𝑡}$
$3𝐴𝑒^{𝑡}+𝐴𝑡𝑒^{𝑡}+8𝐴𝑒^{𝑡}+4𝐴𝑡𝑒^{𝑡}+𝐴𝑒^{𝑡}+𝐴𝑡𝑒^{𝑡}−6𝐴𝑡𝑒^{𝑡}=12𝐴𝑒^{𝑡}$
12𝐴=24
𝐴=2
∴𝑦(𝑡)=$𝑐_{1}𝑒^{𝑡}+𝑐_{2}𝑒^{−2𝑡}+𝑐_{3}𝑒^{−3𝑡}+2𝑡𝑒^{𝑡}$

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