# Toronto Math Forum

## MAT244--2019F => MAT244--Test & Quizzes => Term Test 2 => Topic started by: Victor Ivrii on November 19, 2019, 04:18:46 AM

Title: Problem 2 (morning)
Post by: Victor Ivrii on November 19, 2019, 04:18:46 AM
Consider equation

y'''-2y''+4y'-8y=15\cos (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 (morning)
Post by: Yiheng Bian on November 19, 2019, 04:27:56 AM
No double-dipping
(a):
$$W=ce^{-\int{p(t)}}dt\\ W=ce^{-\int{-2}dt}\\ W=ce^{2t}$$

(b):
We can get:
$$r^3-2r^2+4r-8=0\\ (r-2)(r^2+4)=0\\ r_1=2,r_2=2i,r_3=-2i$$
Therefore:
$$y=c_1e^2t+c_2sin2t+c_3cos2t$$
$$\begin{vmatrix} e^2t & sin2t & cos2t \\ 2e^2t & 2cos2t & -2sin2t \\ 4e^2t & -4sin2t & -4cos2t \end{vmatrix}=e^{2t}({-8(cos2t)^2}-8(sin2t)^2)-sin2t(-8e^{2t}cos2t+8e^{2t}sin2t)+cos2t(-8e^{2t}sin2t-8e^{2t}cos2t)=-8e^{2t}-8e^{2t}=-16e^{2t}$$
compare with (a) we get
$$c=-16$$

(c):
Let
$$y=Acost+Bsint$$
Therefore:
$$y'=-Asint+Bcost\\ y''=-Acost-Bsint\\ y'''=Asint-Bcost$$
Next We take these into equation and get:
$$(-3A-6B)sint+(3B-6A)cost=15cost\\ -3A-6B=0:3B-6A=15\\ A=-2,B=1$$
Finally:
$$y=c_1e^2t+c_2sin2t+c_3cos2t-2cost+sint$$
Title: Re: Problem 2 (morning)
Post by: Lan Cheng on November 19, 2019, 05:58:31 AM
a) We can see the coefficient of y” is -2. Therefore, $W=Ce^{2t}.$

b) First, we should solve for $y'''-2y"+4y'-8y=0.$

Let $r^{3}-2r^{2}+4r-8=0.$

$r_{1}=2,r_{2}=2i,r_{3}=-2i.$

thus, $y_{c}(t)=C_{1}e^{2t}+cos(2t)+sin(2t).$

$W=det\begin{bmatrix}e^{2t} & cos(2t) & sin(2t)\\ 2e^{2t} & -2sin(2t) & 2cos(2t)\\ 4e^{2t} & -4cos(2t) & -4sin(2t) \end{bmatrix}=16e^{2t}.$

Therefore, the answer in b) respond that the differential equation in a) is correct.

c) Let $y_{p}(t)=Acos(t)+Bsin(t).$

$y'=-Asin(t)+Bcos(t).y"=-Acos(t)-Bsin(t).y'''=Asin(t)-Bcos(t).$

Plug in, we get $\begin{cases} -3A-6B=0 & -6A+3B=15.\end{cases}\begin{cases} A=-2 & B=1.\end{cases}$

Therefore, $y(t)=C_{1}e^{2t}+cos(2t)+sin(2t)-2cos(t)+sin(t).$ Two constants missing
Title: Re: Problem 2 (morning)
Post by: Ruojing Chen on November 19, 2019, 06:05:06 AM
$$y'''-2y''+4y'-8y=15cost$$
a) $$p(t)=-2$$
$$w=ce^{-\int p(t)}=ce^{\int2dt}=ce^{2t}$$
b)$$y'''-2y''+4y'-8y=0$$
$$r^3-2r^2+4r-8=0$$
$$(r-2)(r^2+4)=0$$
$$\therefore r=2,\pm2i$$
$$y_c(t)=c_1e^{2t}+c_2cos2t+c_3sin2t$$
$$w(y_1,y_2,y_3)(t)=\left[\begin{matrix} e^{2t} & cos2t & sin2t \\ 2e^{2t} & -2sin2t & 2cos2t \\ 4e^{2t} & -4cos2t & -4sin2t \end{matrix}\right]=16e^{2t}$$
$$\therefore w=ce^{2t}=16e^{2t}, c=16$$
c)Assume $$y_p(t)=Acost+Bsint$$
$$y'=-Asint+Bcost$$$$y''=-Acost-Bsint$$$$y'''=Asint-Bcost$$
By plugging in y,y',y'',y''' into the function
$$Asint-Bcost+2Acost+2Bsint-4Asint+4Bcost-8Acost-8Bsint=15cost$$
$$\left\{ \begin{array}{lr} A+2B-4A-8B=0, & \\ -B+2A+4B-8A=15 \end{array} \right.$$
$$\Rightarrow \left\{ \begin{array}{lr} A+2B=0, & \\ B-2A=5 \end{array} \right.$$
$$\therefore \left\{ \begin{array}{lr} A=-2, & \\ B=1, & \end{array} \right.$$

$$y(t)=c_1e^{2t}+c_2cos2t+c_3sin2t-2cost+sint$$

OK, except LaTeX sucks:

2)  "operators" should be escaped: \cos, \sin, \tan, \ln

$$\boxed{y= -2\cos(t)+\sin(t) + C_1e^{2t} +C_2\cos(2t) +C_3\sin(2t).}$$
Title: Re: Problem 2 (morning)
Post by: Yiran Wang on November 19, 2019, 01:08:10 PM
This is the solution
Title: Re: Problem 2 (morning)
Post by: Yiran Wang on November 19, 2019, 01:08:32 PM
:)
Title: Re: Problem 2 (morning)
Post by: Yiran Wang on November 19, 2019, 01:13:26 PM
:)
Title: Re: Problem 2 (morning)
Post by: Mingdi Xie on November 19, 2019, 02:14:58 PM
I think Lan Cheng miss out $c_2$ and $c_3$ term in front of $\cos2t$ and $\sin2t$