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### Topics - Di Qiu

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1
##### Quiz-6 / Quiz 6 LEC5101
« on: December 07, 2019, 01:40:06 AM »
Question:
a) Find the general solution of the given system of equations.
b) Draw a directions field and a few of the trajectories.  In each of these problems, the coefficient matrix has a zero eigenvalue. As a result, the pattern of trajectories is different from those in the examples in the text.
\begin{align*}
x' =
\begin{pmatrix}
3 & 6 \\
-1 & -2 \\
\end{pmatrix}
x \\
\end{align*}

Solution:
\begin{align*}
det
\begin{pmatrix}
3-\lambda & 6 \\
-1 & -2-\lambda \\
\end{pmatrix}
&=0 \\
(3-\lambda)(-2-\lambda)-(-6) &=0 \\
\lambda^2-\lambda &=0 \\
\lambda(\lambda-1) &=0 \\
\end{align*}
Case 2: $\lambda$=0
\begin{align*}
\begin{pmatrix}
3 & 6 & \vdots & 0\\
-1 & -2 & \vdots & 0\\
\end{pmatrix}
\begin{pmatrix}
3 & 6 & \vdots & 0\\
0 & 0 & \vdots & 0\\
\end{pmatrix}
\begin{pmatrix}
\frac{1}{2} & 1 & \vdots & 0\\
0 & 0 & \vdots & 0\\
\end{pmatrix} \\
\end{align*}
\begin{align*}
\begin{pmatrix}
x1 \\ x2
\end{pmatrix}
=
\begin{pmatrix}
-2 \\ 1
\end{pmatrix}
t
\end{align*}

Case 1: $\lambda$=1
\begin{align*}
\begin{pmatrix}
3-1 & 6 & \vdots & 0\\
-1 & -2-1 & \vdots & 0\\
\end{pmatrix}
\begin{pmatrix}
2 & 6 & \vdots & 0\\
-1 & -3 & \vdots & 0\\
\end{pmatrix}
\begin{pmatrix}
\frac{1}{3} & 1 & \vdots & 0\\
0 & 0 & \vdots & 0\\
\end{pmatrix}
\end{align*}
\begin{align*}
\begin{pmatrix}
x1 \\ x2
\end{pmatrix}
=
\begin{pmatrix}
-3 \\ 1
\end{pmatrix}
t
\end{align*}
Therefore, the general solution is
\begin{align*}
y = c1 *
\begin{pmatrix}
-2 \\ 1
\end{pmatrix}
+ c2 * e^t
\begin{pmatrix}
-3 \\ 1
\end{pmatrix}
\end{align*}

2
##### Quiz-5 / Quiz 5 TUT0402
« on: December 06, 2019, 12:21:46 PM »
Question: Find a particular solution of the given inhomogeneous Euler's equation
$$(1-t)y'' + ty' - y = 2(t-1)^2e^{-t},\quad 0<t<1;\qquad y_1(t)=e^t$$
Solution:
Assume the solution is of the form $y = v(t)*e^t$, then we will have
\begin{align*}
y &= v(t)e^t\\
y' &= v'(t)e^t + v(t)e^t\\
y'' &= v''(t)e^t + 2v'(t)e^t + v(t)e^t
\end{align*}
Plug into the above equation, we have
\begin{align*}
(1-t)y'' + ty' - y &= (1-t)v''(t) + (2-t) v'(t)e^t
\end{align*}
Let r(t) = v'(t), We get
\begin{align*}
(1-t)r'(t)e^t +(2-t)r(t)e^t=2(t-1)^2e^{-t}
\end{align*}
\begin{align*}
r'(t)+\frac{2-t}{1-t}r(t)=2(1-t)e^{-2t}
\end{align*}
Use integrating factor method, we have
\begin{align*}
\mu(t) &= e^{\int\frac{2-t}{1-t}dt} = e^{t-ln(1-t)} = \frac{e^t}{1-t} \\
r(t) &= \frac{\int2(1-t)e^{-2t}*\mu dt}{\mu} = \frac{\int2e^{-t}dt}{e^t/(1-t)}=2(t-1)e^{-2t}
\end{align*}
Integrate r(t), we get
\begin{align*}
v(t) = \int r(t)dt=\int 2(t-1)e^{-2t}=-te^{-2t}+\frac{1}{2}e^{-2t}
\end{align*}
Finally, a particular solution of the given inhomogeneous Euler's equation is
\begin{align*}
y(t)=v(t)e^t=-te^{-t}+\frac{1}{2}e^{-t}
\end{align*}

3
##### Quiz-5 / LEC5101 Quiz5
« on: November 01, 2019, 02:01:58 PM »
Question: $$y'' + 9y' = 9\sec^2(3t), 0<t<\frac{\pi}{6}$$
We first solve the homogeneous part: $$r^2+9=0$$,
where $$r=\pm3i$$,
so $$y_c(t) = c_1\cos{3t}+c_2\sin{3t}$$
Then we use wronskian to solve non-homoogeneous part:
\begin{align*}
w_1 &=
\begin{vmatrix}
0 & \sin{3t} \\
1 & 3\cos{3t} \\
\end{vmatrix} = -\sin{3t} \end{align*}
\begin{align*}
w_2 &=
\begin{vmatrix}
\cos{3t} & 0 \\
-3\sin{3t} & 1 \\
\end{vmatrix} = \cos{3t}
\end{align*}
\begin{align*}
w &=
\begin{vmatrix}
\cos{3t} & \sin{3t} \\
-3\sin{3t} & 3\cos{3t} \\
\end{vmatrix} = 3\cos^2{3t} + 3\sin^2{3t} = 3
\end{align*}
Subsititues above into formula we get:
\begin{align*} Y_p(t) &= \cos{3t} \int{\frac{-\sin{3s}\times 9\sec^2{3s}}{3}ds} + \sin{3t} \int{\frac{\cos{3s}\times9\sec^2{3s}}{3}ds} \\ &= \cos{3t} \cdot -3\int{\sin{3s}\frac{1}{\cos^2{3s}}ds} + \sin{3t} \cdot 3\int{\cos{3s}\frac{1}{\cos^2{3s}}ds}\\ &= \cos{3t} \cdot -3\int{\tan{3s}\cdot\sec{3s}ds} + \sin{3t} \cdot 3\int{\sec{3s}ds}\\ &= \cos{3t} \cdot -\sec{3t} + \sin{3t} \cdot \ln{|\sec{3t}+\tan{3t}|} \\ &= -1 + \sin{3t} \ln{|\sec{3t}+\tan{3t}|} \end{align*}
Finally, $$Y(t) = c_1\cos{3t}+c_2\sin{3t} + \sin{3t} \ln{|\sec{3t}+\tan{3t}|} -1$$

4
##### Quiz-4 / TUT0402 Quiz4
« on: October 18, 2019, 02:00:01 PM »
Find the general solution of y'' - y' - 2y = cosh(2t):
The associated homogenous equation is: \begin{align*} y'' - y' - 2y &= 0 \\ (r-2)(r+1) &= 0 \end{align*}
$$r=2, r=-1$$
Therefore, the complementary function is: $$y_c(t) = c_1e^{-t} + c_2e^{2t}$$
Since $$cosh(t) = \frac{1}{2}(e^t+e^{-t})$$ $$cosh(2t) = \frac{1}{2}(e^{2t}+e^{-2t})$$
we could rewrite the funtion as: $$y'' - y' - 2y = \frac{1}{2}(e^{2t}+e^{-2t})$$
Then we use the method of undetermined coefficients to find the solution for the non-homogeneous function above:
Let $$y_p(t) = Ae^{2t} + Be^{-2t}$$
Then we find $$Ae^{2t}, c_2e^{2t}$$ has the same format, so we times t to get a new equation:
$$y_p(t) = Ate^{2t} + Be^{-2t}$$
Then, $$y_p'(t) = Ae^{2t} + 2Ate^{2t} - 2Be^{-2t}$$ $$y_p''(t) = 2Ae^{2t} + 4Ate^{2t} + 2Ae^{2t} + 4Be^{-2t}$$
Subsititutes back to the equation:
$$2Ae^{2t} + 4Ate^{2t} + 2Ae^{2t} + 4Be^{-2t} -Ae^{2t} - 2Ate^{2t} + 2Be^{-2t} -2Ate^{2t} - 2Be^{-2t} = \frac{1}{2}(e^{2t}+e^{-2t})$$
Then we have: \begin{align*} 2A + 2A - A &= \frac{1}{2} \\ A &= \frac{1}{6} \\ 4B+2B-2B &= \frac{1}{2} \\ B &=\frac{1}{8}\end{align*}
Hence, the particular solution is:
$$y_p(t) = \frac{1}{6}te^{2t} + \frac{1}{8}e^{-2t}$$
Finally, \begin{align*} y(t) &= y_c(t) + y_p(t) \\ &= c_1e^{-t} + c_2e^{2t} + \frac{1}{6}te^{2t} + \frac{1}{8}e^{-2t} \end{align*}

5
##### Quiz-3 / TUT0402 Quiz3
« on: October 11, 2019, 02:01:30 PM »
Find the solution of the given inticial value problem: y'' + 3y' = 0, y(0)=-2, y'(0) = 3.
For problems in form of $$ax^2+bx+c = 0$$, the solution is $$r =\frac{-b\pm{ \sqrt{b^2-4ac}}}{2a}$$
In this case: \begin{align*} r &=\frac{-3\pm{ \sqrt{3^2-4\times 1\times 0}}}{2} \\ &= \frac{-3\pm3}{2} \end{align*}
Therefore we have: $$r = 0, r = -3$$
The genersal solution is: $$y=c_1+c_2e^{-3t}$$
Substitute t=0 into y: $$c_1+c_2 = -2$$
Substitute t=0 into y': $$-3c_2 = 3$$
Solve these two equation we have: $$c_1= -1, c_2 = -1$$
Finally: $$y=-e^{-3t}-1$$

6
##### Quiz-2 / TUT0402 Quiz2
« on: October 04, 2019, 02:06:04 PM »
Question: $$\frac{x}{{(x^2+y^2)}^{\frac{2}{3}}} + \frac{y}{{(x^2+y^2)}^{\frac{3}{2}}} \cdot \frac{dy}{dx} = 0$$
First, we need to find $$M_y(x,y)$$
Where, \begin{align*} M_y(x,y) &= \frac{\partial}{\partial y}M(x,y) \\ &= \frac{\partial}{\partial y}\frac{x}{{(x^2+y^2)}^{\frac{3}{2}}} \\ &= x \cdot \frac{\partial}{\partial y}(x^2+y^2)^{-\frac{3}{2}} \\ &= x \cdot ({-\frac{3}{2}}) \cdot (x^2+y^2)^{-\frac{5}{2}} \cdot 2y \\ &= -\frac{3xy}{(x^2+y^2)^{\frac{5}{2}}} \end{align*}
Then for the $$N_x(x,y)$$
Where, \begin{align*} N_x(x,y) &= \frac{\partial}{\partial x} \\ &= \frac{\partial}{\partial x}\frac{y}{{(x^2+y^2)}^{\frac{3}{2}}} \\ &= y \cdot \frac{\partial}{\partial x}(x^2+y^2)^{-\frac{3}{2}} \\ &= y \cdot ({-\frac{3}{2}}) \cdot (x^2+y^2)^{-\frac{5}{2}} \cdot 2x \\ &= -\frac{3yx}{(x^2+y^2)^{\frac{5}{2}}} \end{align*}
Since $$M_y(x,y) = N_x(x,y)$$ are exact, there exist a Φ, where Φx = M, Φy = N.
Therefore, \begin{align*} \phi &= \int \frac{x}{{(x^2+y^2)}^{\frac{2}{3}}} dx \\ &=\frac{-1}{{(x^2+y^2)}^{\frac{1}{2}}} dx + g(y) \end{align*}
$$\phi_y = \frac{y}{{(x^2+y^2)}^{\frac{2}{3}}} + g'(y)$$
Compare it to N(x,y), we know that g'(y) = 0, so g(y) = constant.
Therefore, $$\phi = \frac{-1}{{(x^2+y^2)}^{\frac{1}{2}}} + C$$
General Solution: $$\frac{-1}{{(x^2+y^2)}^{\frac{1}{2}}} = C$$

7
##### Quiz-1 / TUT0402 Quiz1
« on: September 27, 2019, 06:15:36 PM »
To find the general solution of: $$\frac{dy}{dx}x = (1-y^2)^{\frac{1}{2}}$$
Rearrange: $$\frac{1}{(1-y^2)^{\frac{1}{2}}} dy = \frac{1}{x} dx$$
Take integral on both sides: $$\int \frac{1}{(1-y^2)^{\frac{1}{2}}} dy = \int \frac{1}{x} dx$$
Left-hand side: $$\int (1-y^2)^{\frac{1}{2}} dy = arcsin y$$
Right-hand side: $$\int \frac{1}{x} dx = ln |x|$$
Therefore the general solution: $$arcsin y = ln|x| + C$$
Keep only y on the left-hand side: $$y = sin (ln|x| + C), x \neq 0, y \neq \pm 1$$

8
##### Quiz-1 / TUT0402 Quiz1
« on: September 27, 2019, 02:00:04 PM »
To find the general solution of: $$\frac{dy}{dx}x = (1-y^2)^{\frac{1}{2}}$$
Rearrange: $$\frac{1}{(1-y^2)^{\frac{1}{2}}} dy = \frac{1}{x} dx$$
Take integral on both sides: $$\int \frac{1}{(1-y^2)^{\frac{1}{2}}} dy = \int \frac{1}{x} dx$$
Left-hand side: $$\int (1-y^2)^{\frac{1}{2}} dy = arcsin y$$
Right-hand side: $$\int \frac{1}{x} dx = ln |x|$$
Therefore the general solution: $$arcsin y = ln|x| + C$$
Keep only y on the left-hand side: $$y = sin (ln|x| + C), x \neq 0$$

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