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### Messages - Samarth Agarwal

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1
##### Term Test 2 / Re: TT2A-P2
« on: November 20, 2018, 09:56:01 AM »
Here we can use Abel's Theorem to determine the value of the Wronskian
$$\mbox{Therefore, } W = c e^{2t}, since p(t) = -2$$

for part b, we must find the characteristic equation
$$r^3 - 2r^2 - r + 2 = 0$$
$$(r-1)(r+1)(r-2) = 0$$
$$\mbox{Therefore, } r = 1, -1, 2$$
$$\mbox{Therefore, } y_1(t) = e^{t} y_2(t) = e^{-t} y_3(t) = e^{2t}$$
$$\mbox{Therefore, the Wronskian} W(y_1, y_2, y_3)(t) = \begin{bmatrix} e^{t}&e^{-t}&e^{2t}\\ e^{t}&-e^{-t}&2e^{2t}\\e^{t}&e^{-t}&4e^{2t}\\ \end{bmatrix}$$
$$det(\begin{bmatrix} e^{t}&e^{-t}&e^{2t}\\ e^{t}&-e^{-t}&2e^{2t}\\e^{t}&e^{-t}&4e^{2t}\\ \end{bmatrix}) = -6 e^{2t}$$
This is similiar to the solution found in part a, except that c = -6

for part c,
$$\mbox{Assume} y(t) = Ate^{t}$$
$$\mbox{Therefore, } y'(t) = Ae^{t}(1+t)$$
$$\mbox{Therefore, } y''(t) = Ae^{t}(2+t)$$
$$\mbox{Therefore, } y'''(t) = Ae^{t}(3+t)$$
Plugging this in the original equation
$$Ae^{t}(t + 3 - 4- 2t - 1 - t - 2t) = 8Ae^{t}$$
$$\mbox{Therefore, A = -4}$$
$$\mbox{Therefore, the general equation is } y(t) = c_1e^{t} + c_2e^{-t} + c_3e^{2t} -4te^{t}$$

2
##### Term Test 2 / Re: TT2-P1
« on: November 20, 2018, 09:35:29 AM »
first we change to characteristic equation
$$r^2 + 4 = 0$$
$$\mbox{therefore, } r = \pm 2i$$
$$\mbox{Therefore, the general solution = } y(t) = c_1 \cos2t + c_2 \sin2t$$
The particular solution, upon integration is
$$y_p(t) = -\cos2t(t - \sin(t) \cos(t)) + \sin2t(\log(\cos(t)) - 1/2\cos2t)$$
$$\mbox{Therefore, the general solution is} y(t) = c_1 \cos2t + c_2 \sin2t -\cos2t(t - \sin(t) \cos(t)) + \sin2t(\log(\cos(t)) - 1/2\cos2t)$$
$$y(0) = c_1 = 0$$
upon differentiating y(t) and plugging in t = 0, we get c_2 = 1/2
$$\mbox{Therefore, the general solution is} y(t) = 1/2 \sin2t -\cos2t(t - \sin(t) \cos(t)) + \sin2t(\log(\cos(t)) - 1/2\cos2t)$$

3
##### Term Test 2 / Re: TT2A-P4
« on: November 20, 2018, 08:59:17 AM »
First, try to find the eigenvalues with respect to the parameter
$$A=\begin{bmatrix} 1&2\\ -5&-1\\ \end{bmatrix}$$
$$det(A-rI)=(1-r)(-1-r)+10=0$$
$$r^2 + 9 = 0$$
$$r = \pm 3i$$
The eigenvector is \\ \begin{bmatrix} -2\\ 1-3i \end{bmatrix}
Therefore x_1 =
$$\begin{bmatrix} -2\\ 1-3i \end{bmatrix} (\cos3t + i\sin3t)$$
$$= \begin{bmatrix} -2\cos3t \\ \cos3t + 3\sin3t \end{bmatrix} + i \begin{bmatrix} -2\sin3t\\ -3\cos3t + \sin3t \end{bmatrix}$$
Therefore the general solution
$$x(t) = c_1 \begin{bmatrix} -2\cos3t \\ \cos3t + 3\sin3t \end{bmatrix} + c_2 \begin{bmatrix} -2\sin3t\\ -3\cos3t + \sin3t \end{bmatrix}$$

4
##### Quiz-5 / Re: Q5 TUT 0401
« on: November 04, 2018, 01:58:22 AM »
First, we must change to homogeneous equation
$$y'''(t) + y'(t) = 0$$
Then we must find the characteristic equation
$$r^{3} + r = 0$$
$$r ( r^{2} + 1) = 0$$
$$\mbox{Therefore, } r = 0 \mbox{ or } r = i \mbox{ or } r = -i$$
$$\mbox{Therefore, } y_1(t) = 1 \mbox{ and } y_2(t) = \cos(t) \mbox{ and } y_3(t) = \sin(t)$$
$$\mbox{Therefore, } y_c(t) = c_1 + c_2\cos(t) + c_3\sin(t)$$

$$\mbox{Therefore, the Wronskian = W} (y_1, y_2, y_3)(t) = \begin{vmatrix} 1&\cos(t)&\sin(t)\\ 0&-\sin(t)&\cos(t)\\ 0&-\cos(t)&-\sin(t)\\ \end{vmatrix} = \sin^{2}(t) + \cos^{2}(t) = 1$$

$$\mbox{Now we must find } W_1(y_1, y_2, y_3)$$
$$\mbox{Therefore, } W_1(y_1, y_2, y_3)(t) = \begin{vmatrix} 0&\cos(t)&\sin(t)\\ 0&-\sin(t)&\cos(t)\\ 1&-\cos(t)&-\sin(t)\\ \end{vmatrix} = \cos^{2}(t) + \sin^{2}(t) = 1$$

$$\mbox{Now we must find } W_2(y_1, y_2, y_3)$$
$$\mbox{Therefore, } W_2(y_1, y_2, y_3)(t) = \begin{vmatrix} 1&0&\sin(t)\\ 0&0&\cos(t)\\ 0&1&-\sin(t)\\ \end{vmatrix} = -\cos(t)$$

$$\mbox{Now we must find } W_3(y_1, y_2, y_3)$$
$$\mbox{Therefore, } W_3(y_1, y_2, y_3)(t) = \begin{vmatrix} 1&\cos(t)&0\\ 0&-\sin(t)&0\\ 0&-\cos(t)&1\\ \end{vmatrix} = -\sin(t)$$

$$\mbox{Therefore, }u_1' = \sec(t)$$
$$\mbox{Therefore, }u_2' = \sec(t)(-\cos(t)) = -1$$
$$\mbox{Therefore, }u_3' = \sec(t)(-sin(t)) = \frac{1}{\cos(t)}(-\sin(t)) = -\tan(t)$$

$$\mbox{Therefore, } u_1 = \int_{t_0}^{t} \sec(t) dt = \log(\sec(t) + \tan(t))$$
$$\mbox{Therefore, } u_2 = \int_{t_0}^{t} -1 dt = -t$$
$$\mbox{Therefore, } u_3 = \int_{t_0}^{t} -\tan(t) dt = \log(\cos(t))$$
$$\mbox{Therefore, } y_p(t) = \log(\sec(t) + \tan(t)) + (-t)(\cos(t)) + \log(\cos(t))(\sin(t))$$

$$\mbox{We know that } y(t) = y_c(t) + y_p(t)$$
$$y(t) = c_1 + c_2\cos(t) + c_3\sin(t) + \log\bigl(\sec(t) + \tan(t)\bigr) + (-t)(\cos(t)) + \log(\cos(t))(\sin(t))$$

5
##### Term Test 1 / Re: TT1 Problem 4 (morning)
« on: October 17, 2018, 02:35:44 PM »
First we make the equation into the characteristic equation to find the roots:
$$y''(t) + 4y'(t) + 5y(t) = e^{-2t} + 8sin(t)$$
$$\mbox{Therefore, } r^{2} + 4r + 5 = 0$$
$$\mbox{Therefore, } r = -2 \pm i$$
$$\mbox{Therefore, } y(t) = c_1e^{-2t}cos(t)+ c_2e^{-2t}sin(t)$$
Now we change to non homogeneous equation and try to find the value of A:
$$y''(t) + 4y'(t) + 5y(t) = e^{-2t}$$
Let   $$y(t) = Ae^{-2t}$$
$$\mbox{Therefore, } y'(t) = -2Ae^{-2t} \mbox{ and } y''(t) = 4Ae^{-2t}$$
$$\mbox{Therefore, } 4Ae^{-2t} + 4(-2Ae^{-2t}) + 5Ae^{-2t} = e^{-2t}$$
$$Ae^{-2t} = e^{-2t}$$
$$A = 1$$
Now change to non homogenous equation and find the value of B and C:
$$y''(t) + 4y'(t) + 5y(t) = 8sin(t)$$
Let   $$y(t) = Bcos(t) + Csin(t)$$
$$\mbox{Therefore, } y'(t) = -Bsin(t) + Ccos(t) \mbox{ and } y''(t) = -Bcos(t) -Csin(t)$$
$$\mbox{Therefore, } -Bcos(t) -Csin(t) + 4(-Bsin(t) + Ccos(t)) + 5(Bcos(t) + Csin(t)) = 8sin(t)$$
$$\mbox{Therefore, } -Bcos(t) -Csin(t) + -4Bsin(t) + 4Ccos(t) + 5Bcos(t) + 5Csin(t) = 8sin(t)$$
$$\mbox{Therefore, } 4Bcos(t) + 4Csin(t) - 4Bsin(t) + 4Ccos(t) = 8sin(t)$$
$$sin(t)(4C - 4B) + cos(t)(4B + 4C) = 8sin(t)$$
$$\mbox{Therefore, } 4C - 4B = 8 \mbox{ and } 4B + 4C = 0$$
$$\mbox{Therefore, } B = -1 \mbox{ and } C = 1$$
$$\mbox{Therefore, } y(t) = c_1e^{-2t}cos(t)+c_2e^{-2t}sin(t) + e^{-2t} + sin(t) - cos(t)$$

6
##### Term Test 1 / Re: TT1 Problem 2 (morning)
« on: October 16, 2018, 02:39:54 PM »
Integration of (cosh(x) + 1)/(e^x + 1)^2 wrt dx = -e^(-x)/2 + c
Therefore g = (e^x + 1)(-e^(-x)/2 + c)

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