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Term Test 2 / Re: Problem 4 (noon)
« on: November 19, 2019, 05:46:00 PM »
answer for tt2 question4
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2
Term Test 2 / Re: Problem 3 (noon)
« on: November 19, 2019, 05:42:18 PM »
answer for tt2 question3 b)
3
Term Test 2 / Re: Problem 3 (noon)
« on: November 19, 2019, 05:41:10 PM »
answer for tt2 question3 a)
4
Term Test 2 / Re: Problem 2 (noon)
« on: November 19, 2019, 05:37:38 PM »
answer for tt2 question2
6
Quiz-5 / quiz5 tut0401
« on: November 01, 2019, 01:33:15 PM »
Solve $y'' + 4y = 3csc2t$, $0 < t<\frac{\pi}{2}$
For solution to homogeneous equation $y'' + 4y = 0$, the characteristic polynomial is
\begin{align*}
r^2 + 4 &= 0\\
r_1 &= 2i\quad r_2 = -2i
\end{align*}
therefore, solution to homogeneous equation
\begin{align*}
y_c &= C_1e^{\lambda t}cos\mu t + C_2e^{\lambda t }sin\mu t\\
&= C_1cos2t + C_2sin2t
\end{align*}
For solution to $y'' + 4y = 3csc(2t)$
, since $p(t) = 0$, $q(t) = 4$, $g(t) = 3csc(2t)$ are both continuous on $0< t< \frac{\pi}{2}$
\begin{equation*}
W = \det \begin{bmatrix}
y_1 & y_1' \\ y_2 & y_2'
\end{bmatrix} = \begin{bmatrix}
cos2t & sin2t' \\ -2sin2t & 2cos2t'
\end{bmatrix} = 2
\end{equation*}
Thus, $y_1(t)$ and $y_2(t)$ form a fundamental set of solutions.
Therefore
\begin{align*}
u_1(t) &= -\int \frac{y_2(t)g(t)}{W[y_1, y_2](t)}\\
&= -\int\frac{(sin2t)(3csc2t)}{2}dt\\
&= -\int\frac{3}{2}dt\\
&= -\frac{3}{2}t
\end{align*}
\begin{align*}
u_2(t) &= \int\frac{y_1(t)g(t)}{W[y_1, y_2](t)}\\
&= \int\frac{(cos2t)(3csc2t)}{2}dt\\
&= \frac{3}{2}\int\frac{\cos{2t}}{\sin{2t}}dt\\
&= \frac{3}{2}\int \cot{2t}dt\\
&= \frac{3}{4}\ln{|\sin{2t}|}
\end{align*}
Therefore, the particular solution is $y_p(t) = u_1(t)y_1(t) + u_2(t)y_2(t)$, we get
\begin{align*}
y_p &= \cos{2t}(-\frac{3}{2}t) + \sin{2t}(\frac{3}{4}\ln{|\sin{2t}|})\\
&= \frac{3}{4}\sin{2t}\ln{|\sin{2t}|} - \frac{3}{2}t\cos{2t}
\end{align*}
Therefore the general solution is
\begin{equation*}
C_1cos2t + C_2sin2t + \frac{3}{4}\sin{2t}\ln{|\sin{2t}|} - \frac{3}{2}t\cos{2t}
\end{equation*}
For solution to homogeneous equation $y'' + 4y = 0$, the characteristic polynomial is
\begin{align*}
r^2 + 4 &= 0\\
r_1 &= 2i\quad r_2 = -2i
\end{align*}
therefore, solution to homogeneous equation
\begin{align*}
y_c &= C_1e^{\lambda t}cos\mu t + C_2e^{\lambda t }sin\mu t\\
&= C_1cos2t + C_2sin2t
\end{align*}
For solution to $y'' + 4y = 3csc(2t)$
, since $p(t) = 0$, $q(t) = 4$, $g(t) = 3csc(2t)$ are both continuous on $0< t< \frac{\pi}{2}$
\begin{equation*}
W = \det \begin{bmatrix}
y_1 & y_1' \\ y_2 & y_2'
\end{bmatrix} = \begin{bmatrix}
cos2t & sin2t' \\ -2sin2t & 2cos2t'
\end{bmatrix} = 2
\end{equation*}
Thus, $y_1(t)$ and $y_2(t)$ form a fundamental set of solutions.
Therefore
\begin{align*}
u_1(t) &= -\int \frac{y_2(t)g(t)}{W[y_1, y_2](t)}\\
&= -\int\frac{(sin2t)(3csc2t)}{2}dt\\
&= -\int\frac{3}{2}dt\\
&= -\frac{3}{2}t
\end{align*}
\begin{align*}
u_2(t) &= \int\frac{y_1(t)g(t)}{W[y_1, y_2](t)}\\
&= \int\frac{(cos2t)(3csc2t)}{2}dt\\
&= \frac{3}{2}\int\frac{\cos{2t}}{\sin{2t}}dt\\
&= \frac{3}{2}\int \cot{2t}dt\\
&= \frac{3}{4}\ln{|\sin{2t}|}
\end{align*}
Therefore, the particular solution is $y_p(t) = u_1(t)y_1(t) + u_2(t)y_2(t)$, we get
\begin{align*}
y_p &= \cos{2t}(-\frac{3}{2}t) + \sin{2t}(\frac{3}{4}\ln{|\sin{2t}|})\\
&= \frac{3}{4}\sin{2t}\ln{|\sin{2t}|} - \frac{3}{2}t\cos{2t}
\end{align*}
Therefore the general solution is
\begin{equation*}
C_1cos2t + C_2sin2t + \frac{3}{4}\sin{2t}\ln{|\sin{2t}|} - \frac{3}{2}t\cos{2t}
\end{equation*}
7
Quiz-4 / QUIZ4 TUT0401
« on: October 18, 2019, 02:37:19 PM »
Solve $y'' + 2y' + 5y = 3sin(2t)$
\begin{align*}
r^2 + 2r + 5 &= 0\\
r^2 + 2r + 1 &= -4\\
(r+1)^2 &= -4\\
r + 1 &= \pm 2i\\
r &= -1 \pm 2i
\end{align*}
For solution to homogeneous equation
\begin{align*}
y_c &= C_1e^{\lambda t}cos\mu t + C_2e^{\lambda t }sin\mu t\\
&= C_1e^{-t}cos2t + C_2e^{-t}sin2t
\end{align*}
For solution to $y'' + 2y' + 5y = 3sin(2t)$, guess $y_p = Asin(2t) + Bcos(2t)$
\begin{align*}
y_p' &= 2Acos(2t) - 2Bsin(2t)\\
y_p'' &= -4Asin(2t) - 4cos(2t)
\end{align*}
Then
\begin{align*}
y'' + 2y' + 5y &= -4Asin(2t) - 4cos(2t) + 4Acos(2t) - 4Bsin(2t) + 5Asin(2t) + 5Bcos(2t)\\
&= (A - 4B)sin(2t) + (B + 4A)cos(2t)\\
&= 3sin(2t)
\end{align*}
Therefore
\begin{align*}
&A - 4B = 3 \quad and \quad B +4A = 0\\
&A = \frac{3}{17}\quad and \quad B = - \frac{12}{17}
\end{align*}
The solution to $y'' + 2y' + 5y = 3sin(2t)$ is $y_p = \frac{3}{17}sin(2t) - \frac{12}{17}cos(2t)$
The general solution is
\begin{equation*}
y = C_1e^{-t}cos2t + C_2e^{-t}sin2t + \frac{3}{17}sin(2t) - \frac{12}{17}cos(2t)
\end{equation*}
\begin{align*}
r^2 + 2r + 5 &= 0\\
r^2 + 2r + 1 &= -4\\
(r+1)^2 &= -4\\
r + 1 &= \pm 2i\\
r &= -1 \pm 2i
\end{align*}
For solution to homogeneous equation
\begin{align*}
y_c &= C_1e^{\lambda t}cos\mu t + C_2e^{\lambda t }sin\mu t\\
&= C_1e^{-t}cos2t + C_2e^{-t}sin2t
\end{align*}
For solution to $y'' + 2y' + 5y = 3sin(2t)$, guess $y_p = Asin(2t) + Bcos(2t)$
\begin{align*}
y_p' &= 2Acos(2t) - 2Bsin(2t)\\
y_p'' &= -4Asin(2t) - 4cos(2t)
\end{align*}
Then
\begin{align*}
y'' + 2y' + 5y &= -4Asin(2t) - 4cos(2t) + 4Acos(2t) - 4Bsin(2t) + 5Asin(2t) + 5Bcos(2t)\\
&= (A - 4B)sin(2t) + (B + 4A)cos(2t)\\
&= 3sin(2t)
\end{align*}
Therefore
\begin{align*}
&A - 4B = 3 \quad and \quad B +4A = 0\\
&A = \frac{3}{17}\quad and \quad B = - \frac{12}{17}
\end{align*}
The solution to $y'' + 2y' + 5y = 3sin(2t)$ is $y_p = \frac{3}{17}sin(2t) - \frac{12}{17}cos(2t)$
The general solution is
\begin{equation*}
y = C_1e^{-t}cos2t + C_2e^{-t}sin2t + \frac{3}{17}sin(2t) - \frac{12}{17}cos(2t)
\end{equation*}
8
Quiz-3 / tut0401 Quiz3
« on: October 11, 2019, 01:59:16 PM »
Find the Wronskian of the given pair of functions:
$$
cos^2(x),\,1+cos(2x)
$$
$$
W=
\begin{vmatrix}
cos^2(x)&1+cos(2x)\\
-2cos(x)sin(x)&-2sin(2x)\\
\end{vmatrix}\\
=
\begin{vmatrix}
cos^2(x)&1+cos(2x)\\
-sin(2x)&-2sin(2x)\\
\end{vmatrix}\\
cos(2x)=2cos^2(x)-1\\
sin(2x)=2sin(x)cos(x)\\
=-2cos^2(x)sin(2x)+sin(2x)+sin(2x)cos(2x)\\
=-sin(2x)[2cos^2(x)-1-cos(2x)]\\
=-sin(2x)[cos(2x)-cos(2x)]\\
=-sin(2x)\times0\\
=0\\
\
$$
$$
cos^2(x),\,1+cos(2x)
$$
$$
W=
\begin{vmatrix}
cos^2(x)&1+cos(2x)\\
-2cos(x)sin(x)&-2sin(2x)\\
\end{vmatrix}\\
=
\begin{vmatrix}
cos^2(x)&1+cos(2x)\\
-sin(2x)&-2sin(2x)\\
\end{vmatrix}\\
cos(2x)=2cos^2(x)-1\\
sin(2x)=2sin(x)cos(x)\\
=-2cos^2(x)sin(2x)+sin(2x)+sin(2x)cos(2x)\\
=-sin(2x)[2cos^2(x)-1-cos(2x)]\\
=-sin(2x)[cos(2x)-cos(2x)]\\
=-sin(2x)\times0\\
=0\\
\
$$
9
Quiz-2 / QUIZ2 TUT0401
« on: October 04, 2019, 02:00:07 PM »
Solve \begin{align*}
(2xy^2 + 2y) + (2x^2y + 2x)y' &= 0\\
\implies (2xy^2 + 2y)dx + (2x^2y + 2x)dy &= 0
\end{align*}
Let $M = 2xy^2 + 2y$ and $N = 2x^2y + 2x$
\begin{equation}
M_y = \frac{d M}{dy} = 4xy + 2x\quad\quad\quad N_x = \frac{d N}{dy} = 4xy + 2x
\end{equation}
Since $M_y = N_x$, the equation is exact so $\exists \psi(x, y)\quad s.t \quad \frac{\partial \psi}{\partial x} = M$ and $\frac{\partial \psi}{\partial y} = N$
Therefore
\begin{align*}
\psi(x, y) &= \int M dx \\
&= \int 2x^2y + 2y dx\\
&= x^2y^2 + 2xy + h(y)
\end{align*}
Since
\begin{align*}
\frac{\partial \psi}{\psi y} &= 2x^2y + 2x +h'(y)\\
\implies h'(y) &= 0\\
\implies h(y) &= C
\end{align*}
Therefore, the solution is $x^2y^2 + 2xy = C$
(2xy^2 + 2y) + (2x^2y + 2x)y' &= 0\\
\implies (2xy^2 + 2y)dx + (2x^2y + 2x)dy &= 0
\end{align*}
Let $M = 2xy^2 + 2y$ and $N = 2x^2y + 2x$
\begin{equation}
M_y = \frac{d M}{dy} = 4xy + 2x\quad\quad\quad N_x = \frac{d N}{dy} = 4xy + 2x
\end{equation}
Since $M_y = N_x$, the equation is exact so $\exists \psi(x, y)\quad s.t \quad \frac{\partial \psi}{\partial x} = M$ and $\frac{\partial \psi}{\partial y} = N$
Therefore
\begin{align*}
\psi(x, y) &= \int M dx \\
&= \int 2x^2y + 2y dx\\
&= x^2y^2 + 2xy + h(y)
\end{align*}
Since
\begin{align*}
\frac{\partial \psi}{\psi y} &= 2x^2y + 2x +h'(y)\\
\implies h'(y) &= 0\\
\implies h(y) &= C
\end{align*}
Therefore, the solution is $x^2y^2 + 2xy = C$
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