# Toronto Math Forum

## MAT244--2019F => MAT244--Test & Quizzes => Term Test 2 => Topic started by: Victor Ivrii on November 19, 2019, 04:17:26 AM

Title: Problem 2 (main sitting)
Post by: Victor Ivrii on November 19, 2019, 04:17:26 AM
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}).
Title: Re: Problem 2 (main sitting)
Post by: Yiheng Bian on November 19, 2019, 04:29:25 AM
(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.
Title: Re: Problem 2 (main sitting)
Post by: xuanzhong on November 19, 2019, 05:03:39 AM
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, $y(t)=c_1e^t+c_2e^{-2t}+c_3e^{-3t}+2te^t$
Title: Re: Problem 2 (main sitting)
Post by: Yuying Chen on November 19, 2019, 05:42:54 AM
$\text(a)\\$
$W=ce^{-\int p(t)dt}=ce^{-\int 4dt}=ce^{-4t}\\$
$\text(b)\\$
$r^3+4r^2+r-6=0\\$
$(r-1)(r^2+5r+6)=0\\$
$r=1, r=-2, r=-3\\$
$\therefore y(t)=c_1e^t+c_2e^{-2t}+c_3e^{-3t}\\$
$W=\begin{vmatrix} e^t & e^{-2t} & e^{-3t}\\ e^t & -2e^{-2t} & -3e^{-3t}\\ e^t & 4e^{-2t} & 9e^{-3t}\\ \end{vmatrix}=-12e^{-4t}\\$
$\text{$\therefore c=-12$compare with (a)}\\$
$\text(c)\\$
$y_p(t)=Ate^t\\$
$y_p^{\prime}(t)=Ae^t+Ate^t\\$
$y_p^{\prime\prime}(t)=2Ae^t+Ate^t\\$
$y_p^{\prime\prime\prime}(t)=3Ae^t+Ate^t\\$
$3Ae^t+Ate^t+8Ae^t+4Ate^t+Ae^t+Ate^t-6Ate^t=12Ae^t\\$
$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad 12A=24\\$
$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad A=2$
$\therefore y(t)=c_1e^t+c_2e^{-2t}+c_3e^{-3t}+2te^t$
Title: Re: Problem 2 (main sitting)
Post by: Xinyu Jing on November 19, 2019, 06:30:34 AM
(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)=$𝑐_{1}𝑒^{𝑡}+𝑐_{2}𝑒^{−2𝑡}+𝑐_{3}𝑒^{−3𝑡}+2𝑡𝑒^{𝑡}$
Title: Re: Problem 2 (main sitting)
Post by: Ruodan Chen on November 19, 2019, 10:32:48 AM
2) $y'''+4y''+y'-6y=24e^{t}$

a)

Since the coefficient of y'' is 4,

Then, Wronskain is

$w=ce^{-\int4dt}=ce^{-4t}$

b)

Use homogeneous equation to find fundamental solutions,

$y'''+4y''+y'-6y=24e^{t}$

Then $r^{3}+4r^{2}+r-6=0$

Then $r=1, r=-2, r=-3$

Then the solution is $y_{c}(t)=c_{1}e^{t}+c_{2}e^{-2t}+c_{3}e^{-3t}$

$w = \begin{array}{ccc} e^{t} & e^{-2t} & e^{-3t}\\ e^{t} & -2e^{-2t} & -3e^{-3t}\\ e^{t} & 4e^{-2t} & 9e^{-3t} \end{array}=-12e^{-4t}$

$w = -12e^{-4t} =ce^{-4t}$

This is consistent with what we get in part (a) with $c=-12$

c)

Use under determined coefficients method to find the general equation for y(t)

$y'''+4y''+y'-6y=24e^{t}$

$y_{p}(t)=Ate^{t}$

$y'_{p}(t)=Ae^{t}+Ate^{t}$

$y''_{p}(t)=2Ae^{t}+Ate^{t}$

$y'''_{p}(t)=3Ae^{t}+Ate^{t}$

Plug into the equation,

$3Ae^{t}+Ate^{t}+4(2Ae^{t}+Ate^{t})+Ae^{t}+Ate^{t}-6Ate^{t}=12Ae^{t}=24e^{t}$

We get A=2

$y_{p}(t)=2te^{t}$

Thus, $y(t)=c_{1}e^{t}+c_{2}e^{-2t}+c_{3}e^{-3t}+2te^{t}$
Title: Re: Problem 2 (main sitting)
Post by: nayan on November 19, 2019, 01:59:47 PM
$\text(a)\\$

$\text{Using abels theorem,} \\ W=ce^{-\int p(t)dt}=ce^{-\int 4dt}=ce^{-4t}\\$
$\text(b)\\$
$\text{We have the following characteristic polynomial,}\\$
$r^3+4r^2+r-6=0\\$
$(r-1)(r^2+5r+6)=0\\$
$r=1, r=-2, r=-3\\$
$\therefore y(t)=c_1e^t+c_2e^{-2t}+c_3e^{-3t}\\$
$W=\begin{vmatrix} e^t & e^{-2t} & e^{-3t}\\ e^t & -2e^{-2t} & -3e^{-3t}\\ e^t & 4e^{-2t} & 9e^{-3t}\\ \end{vmatrix}=-12e^{-4t}\\$
$\text{$\therefore c=-12$}\\$
$\text(c)\\$
$y_p(t)=Ate^t\\$
$y_p^{\prime}(t)=Ae^t+Ate^t\\$
$y_p^{\prime\prime}(t)=2Ae^t+Ate^t\\$
$y_p^{\prime\prime\prime}(t)=3Ae^t+Ate^t\\$
$3Ae^t+Ate^t+8Ae^t+4Ate^t+Ae^t+Ate^t-6Ate^t=12Ae^t\\$
$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad 12A=24\\$
$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad A=2$
$\text{We finally have the general solution,}\\ y(t)=c_1e^t+c_2e^{-2t}+c_3e^{-3t}+2te^t$
Title: Re: Problem 2 (main sitting)
Post by: Mingdi Xie on November 19, 2019, 08:18:51 PM
This is my solution :)