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### Messages - ChenYang Li

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1
##### Quiz-5 / LEC 0101 Quiz 5
« on: November 01, 2019, 02:03:31 PM »
First verify the two solutions, then find a particular solution

t^2y'' - t(t+2)y'+(t+2)y = 2t^3 \qquad t>0\\
y_1 = t \qquad \qquad y_2 = te^t

First, verify the two solutions:

y_1 = t \qquad y'_1 = 1 \qquad y''_1 = 0\\
y_2 = te^t \qquad y'_2 = (1+t)e^t \qquad y''_2 = (2+t)e^t\\
For \quad y_1 \\
t^2(0) - t(t+2)(1)+(t+2)t = 0\\
0-t(t+2)+t(t+2)=0\\
0=0\\
Thus, \quad y_1 \quad is \quad solution\quad of \quad euqation\\
Then, \quad for\quad y_2\\
t^2(2+t)e^t - t(t+2)(1+t)e^t+(t+2)te^t =0\\
t^2(2+t) - t(t+2)(1+t)+(t+2)t =0\\
t^2-t(1+t)+t =0\\
t^2-t^2-t+t=0\\
0=0
y_2 \quad is \quad also\quad a \quad soltuion

Then, let's calculate the particular solution

\begin{split}
W & =
\begin{vmatrix}
y_1 & y_2 \\
y'_1 & y'_2 \\
\end{vmatrix}\\
& =
\begin{vmatrix}
t & te^t \\
1 & (1+t)e^t \\
\end{vmatrix}\\
& = t(1+t)e^t - te^t\\
& = t^2e^t
\end{split}\\
\begin{split}
W_1 & =
\begin{vmatrix}
0 & te^t \\
1 & (1+t)e^t \\
\end{vmatrix}\\
& = 0-te^t\\
& = -te^t
\end{split}\\
\begin{split}
W_2 & =
\begin{vmatrix}
t & 0 \\
1 & 1 \\
\end{vmatrix}\\
& = t
\end{split}

Next,

\begin{split}
u_1 & = \int \frac{-te^t(2t^3)}{t^2e^t} dt\\
& = -\int 2t^2dt\\
& = -\frac{2}{3}t^3
\end{split}\\
\begin{split}
u_2 & = \int \frac{t(2t^3)}{t^2e^t} dt\\
& = \int \frac{2t^2}{e^t}\\
& = -2t^2e^{-t}-4te^{-t}-4e^{-t}
\end{split}\\

Finally,

\begin{split}
Y_p & = y_1(u_1) + y_2(u_2)\\
& = -\frac{2}{3}t^4-2t^3+4t^2-4t
\end{split}\\

2
##### Term Test 1 / Re: Problem 4 (main)
« on: October 23, 2019, 07:03:40 AM »
TT1 Question 4 (Main)
a).
Solve

y''-6y'+10y=2e^{3x}+39cosx

First solve for

r^2-6r+10=0\\
\begin{split}
r & =\frac{-(-6)\pm \sqrt{(-6)^2-4(1)(10)}}{2(1)}\\
& = \frac{6\pm \sqrt{-4}}{2}\\
& = 3\pm i\\
\end{split}\\
y_c=c_1e^{3x}cosx+c_2e^{3x}sinx

Then, we solve

y''-6y'+10y=2e^{3x}\\
y_{p1}=Ae^{3x} \qquad y_{p1}'=3Ae^{3x} \qquad y_{p1}''=9Ae^{3x}\\
9Ae^{3x}-6(3Ae^{3x})+10(Ae^{3x})=2e^{3x}\\
9A-18A+10A=2\\
A=2\\
y_{p1}=2e^{3x}

Next, solve

y''-6y'+10y=39cosx\\
y_{p2}=Bcosx+Csinx \qquad y_{p2}'=-Bsinx+Ccosx \qquad y_{p2}''=-Bcosx-Csinx\\
(-Bcosx-Csinx)-6(-Bsinx+Ccosx)+10(Bcosx+Csinx)=39cosx\\
-Bcosx-6Ccosx+10Bcosx=39cosx\\
-Csinx+6Bsinx+10Csinx=0sinx\\
9B-6C=39\\
6B+9C=0\\
B=3 \qquad C=-2\\
y_{p2}=3cosx-2sinx

Now, we add to the general form:

\begin{split}
y& =y_c+y_{p1}+y_{p2}\\
& =c_1e^{3x}cosx+c_2e^{3x}sinx+2e^{3x}+3cosx-2sinx
\end{split}\\

b).
Input:  y(0)=0  y'(0)=0
Get the final solution

\begin{split}
y(0)& =c_1e^0cos0+c_2e^{0}sin0+2e^{0}+3cos0-2sin0\\
& =c_1(1)+c_2(0)+2(1)+3(1)-2(0)\\
& =c_1+5\\
& =0\\
\end{split}\\
c_1=-5\\
\quad \\
y'=3c_1e^{3x}cosx-c_1e^{3x}sinx+3c_2e^{3x}sinx+c_2e^{3x}cosx+6e^{3x}-3sinx-2cosx\\
\begin{split}
y'(0)& =3c_1e^{0}(1)-c_1e^{0}(0)+3c_2e^{0}(0)+c_2e^{0}(1)+6e^{0}-3(0)-2(1)\\
& =3c_1+c_2+6-2\\
& =3(-5)+c_2+4\\
& =c_2-11\\
& =0
\end{split}\\
c_2=11\\

Now we know the two constant c, we plug them into the solution, get:

y=-5e^{3x}cosx+11e^{3x}sinx+2e^{3x}+3cosx-2sinx

OK. V.I.

3
##### Quiz-4 / TUT0201 Quiz4
« on: October 18, 2019, 02:00:09 PM »
Find the general solution of
y''+12y'+9y=0

Solve:

4r^2+12r+9r=0\\
(2r+3)^2=0\\
r_1=r_2=-\frac{3}{2}\\
Thus, \qquad \qquad \qquad \qquad \qquad\\
y=c_1e^{-\frac{3t}{2}}+c_2te^{-\frac{3t}{2}}

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