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### Messages - 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
##### Term Test 1 / Re: Problem 3 (afternoon)
« on: October 23, 2019, 07:46:30 AM »
Sorry,but there is a mistake at the first step, where $$(r-2)(r-3)=0$$, so $$r=2, r=3$$
a) 1. let $y"-5y'+6y=0.$

$r^{2}-5r+6=0.$

$\begin{cases} r_{1}=-2 & r_{2}=-3.\end{cases}$

Thus, $y_{c}(x)=C_{1}e^{-2x}+C_{2}e^{-3x}.$

2.let $y"-5y'+6y=52cos(2x).$

let $y_{p}(x)=Acos(2x)+Bsin(2x). y'=-2Asin(2x)+2Bcos(2x),y"=-4Acos(2x)-4Bsin(2x).$

$-4Acos(2x)-4Bsin(2x)+10Asin(2x)-10Bcos(2x)+6Acos(2x)+6Bsin(2x)=52cos(2x).$

$\begin{cases} -4A-10B+6A=52 & -4B+10A+6B=0\end{cases}.$

$A=1,B=-5.$

$y_{p}(x)=cos(2x)-5sin(2x).$

Therefore, $y(x)=C_{1}e^{-2x}+C_{2}e^{-3x}+cos(2x)-5sin(2x).$

b) $y’(x)=-2C_{1}e^{-2x}-3C_{2}e^{-3x}-2sin(2x)-10cos(2x).$

$\begin{cases} C_{1}+C_{2}+1=0 & -2C_{1}-3C_{2}-10=0\end{cases}.$

$C_{1}=7,C_{2}=-8.$

Therefore, $y(x)=7e^{-2x}-8e^{-3x}+cos(2x)-5sin(2x).$

5
##### Term Test 1 / Re: Problem 3 (afternoon)
« on: October 23, 2019, 07:27:06 AM »
a)
$$(r-2)(r-3)=0$$
$$r=2, r=3$$
$$y_c = c_1e^{2x}+c_2e^{3x}$$
$$y_{p}=A\sin{2x}+B\cos{2x}$$
$$y'_{p}=2A\cos{2x}-2B\sin{2x}$$
$$y''_{p2}=-4A\sin{2x}-4B\cos{2x}$$
Therefore, $$-4A\sin{2x}-4B\cos{2x}-10A\cos{2x}+10B\sin{2x}+6A\sin{2x}+6B\cos{2x}=52\cos{2x}$$
$$A=-5, B=1$$
$$y_{p}=c_1e^{-x}+c_2e^{6x}$$
$$y=c_1e^{2x}+c_2e^{3x}+\cos{2x}-5\sin{2x}$$
b)
$$y'=2c_1e^{2x}+3c_2e^{3x}-2\sin{2x}-10\cos{2x}$$
substitutes $$y(0)=0, y'(0)=0$$
$$c_1+c_2+1=0$$
$$2c_1+3c_2-10=0$$
$$c_2=12, c_1=-13$$
$$y=-13e^{2x}+12e^{3x}+\cos{2x}-5\sin{2x}$$

6
##### Term Test 1 / Re: Problem 4 (afternoon)
« on: October 23, 2019, 06:51:25 AM »
a)
$$r = \frac{-2\pm\sqrt{4-4\cdot17}}{2} = -1\pm4i$$
$$y_c = c_1e^{-x}cos4x+c_2e^{-x}sin4x$$
$$y_{p1}=Ae^x$$
$$y'_{p1}=Ae^x$$
$$y''_{p1}=Ae^x$$
Therefore, $$Ae^x+2Ae^x+17Ae^x = 40e^x$$
$$A=2$$
$$y_{p1}=2e^x$$
$$y_{p2}=B\sin{4x}+C\cos{4x}$$
$$y'_{p2}=4B\cos{4x}-4C\sin{4x}$$
$$y''_{p2}=-16B\sin{4x}-16C\cos{4x}$$
Therefore, $$-16B\sin{4x}-16C\cos{4x}+ 8B\cos{4x}-8C\sin{4x}+17B\sin{4x}+17C\cos{4x}=130\sin{4x}$$
$$B=2, C=-16$$
$$y_{p2}=2\sin{4x}-16\cos{4x}$$
Then we have:
$$y=c_1e^{-x}cos4x+c_2e^{-x}sin4x+2e^x+2\sin{4x}-16\cos{4x}$$
b)
substitutes $$y(0)=0, y'(0)=0$$ into y and y':
$$c_1+2-16=0$$
$$-c_1+4c_2+2+8=0$$
$$c_1=14, c_2=1$$
$$y=14e^{-x}cos4x+e^{-x}sin4x+2e^x+2\sin{4x}-16\cos{4x}$$

OK. V.I.

7
##### 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*}

8
##### 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: $$y=-e^{-3t}-1$$

9
##### 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: $$\frac{-1}{{(x^2+y^2)}^{\frac{1}{2}}} = C$$

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