Show Posts

This section allows you to view all posts made by this member. Note that you can only see posts made in areas you currently have access to.


Messages - Yiheng Bian

Pages: [1] 2
1
Quiz 3 / TUT5101 QUIZ3
« on: February 06, 2020, 12:34:14 AM »
$$
\int_{\Upsilon}e^z dz
$$
$$
\text{line from 0 to } z_{0}$$
Therefore
$$
r(t)= tz_{0}
$$
$$
r'(t) = z_{0} (0\leq t \leq 1)
$$
And since
$$
f(z)=e^z
$$
$$
f(r(t))= e^{z_{0}}
$$
So
$$
\int_{\Upsilon}e^z dz =  \int_{0}^{1}f(r(t))r'(t)dt = \int_{0}^{1}e^{tz_{0}}z_{0} dt = z_{0}(\frac{1}{z_{0}}e^{z_{0}}- \frac{1}{z_{0}})= e^{z_{0}} - 1
$$

2
Quiz 2 / Re: TUT5101 QUIZ2
« on: January 31, 2020, 11:31:11 PM »
Ok, I got it. Thank you sir

3
Quiz 2 / Re: TUT5101 QUIZ2
« on: January 30, 2020, 09:25:16 PM »
Sorry sir, , could you point it out?

4
Quiz 2 / TUT5101 QUIZ2
« on: January 30, 2020, 12:42:13 AM »
Question:
$$
h(z)=\frac{Log(z)}{z}
$$
We know when
$$
 z \rightarrow \infty
$$
$$
Log(z) \rightarrow \infty
$$
So we can use L'Hopital rules and get
$$
\frac{1}{z}
$$
So
$$
\text{limit is zero}
$$


5
Quiz 1 / Re: TUT5101
« on: January 26, 2020, 04:25:25 PM »
I have modified it sir.

6
Quiz 1 / TUT5101
« on: January 26, 2020, 01:14:20 AM »
Find all solutions of the given equation:
$$
z^8 =1
$$
We know
$$
z=r(\cos\theta +I\sin\theta)
$$
$$
z^8=r^8(\cos8\theta +I\sin8\theta)
$$
Then
$$
1=1+0i=1(\cos2k\pi + I\sin2k\pi)
$$
So we have
$$
r^8(\cos8\theta +I\sin8\theta) = 1(\cos2k\pi + I\sin2k\pi)
$$
Therefore we get
$$
r=1
$$
$$
8\theta=2k\pi
$$
$$
\theta= \frac{k\pi}{4}
$$
So when
$$
k=0, \theta=0,z=1(\cos0+i\sin0)=1
$$
$$
k=1, \theta=\frac{\pi}{4},z=1(\cos(\frac{\pi}{4})+I\sin{\frac{\pi}{4}})
$$
$$
k=2, \theta=\frac{\pi}{2},z=1(\cos(\frac{\pi}{2})+i\sin{\frac{\pi}{2}})
$$
$$
k=3, \theta=\frac{3\pi}{4},z=1(\cos(\frac{3\pi}{4})+i\sin{\frac{3\pi}{4}})
$$
$$
k=4, \theta=\pi,z=1(\cos\pi+i\sin\pi)
$$
$$
k=5, \theta=\frac{5\pi}{4},z=1(\cos(\frac{5\pi}{4})+i\sin{\frac{5\pi}{4}})
$$
$$
k=6, \theta=\frac{3\pi}{2},z=1(\cos(\frac{3\pi}{2})+i\sin{\frac{3\pi}{2}})
$$
$$
k=7, \theta=\frac{7\pi}{4},z=1(\cos(\frac{7\pi}{4})+i\sin{\frac{7\pi}{4}})
$$

7
Term Test 2 / Re: Problem 1 (noon)
« on: November 19, 2019, 04:33:17 AM »
No double-dipping

(a):
We can solve homo firstly:
$$
r^2-3r+2=0\\
(r-2)(r-1)=0\\
r_1=2,r_2=1
$$
Therefore:
$$
y=c_1e^{2t}+c_2e^t
$$
So we can get:
$$
W=\begin{vmatrix}
e^{2t} & e^t \\
2e^{2t} & e^t
\end{vmatrix}=-e^{3t}\\
W_1=\begin{vmatrix}
0 & e^{t} \\
1 & e^{t}
\end{vmatrix}=-e^{t}\\
W_2=\begin{vmatrix}
e^{2t} & 0 \\
2e^{2t} & 1
\end{vmatrix}=e^{2t}
$$
So we can get:
$$
Y(t)=e^{2t}\int{\frac{-e^{s}*\frac{e^{3s}}{e^{s2}+1}}{-e^{3s}}}ds + e^{t}\int{\frac{e^{2s}*\frac{e^{3s}}{e^{s2}+1}}{-e^{3s}}}ds\\
Y(t)=e^{2t}\int{\frac{e^{s}}{e^{2s}+1}}ds - e^{t}\int{\frac{e^{2s}}{e^{2s}+1}}ds\\
Y(t)=e^{2t}*arctan(e^t) - 0.5e^{t}*ln(e^{2t}+1)
$$
Finally:
$$
y(t)=c_1e^{2t}+c_2e^t+e^{2t}*arctan(e^t) - 0.5e^{t}*ln(e^{2t}+1)
$$




(b):
So we can get y'(t):
$$
y'=2c_1e^{2t}+c_2e^t+2e^{2t}arctan(e^t)+e^{2t}*\frac{e^t}{e^{2t}+1}-0.5e^t*ln(e^{2t}+1)-0.5e^t*\frac{e^{2t}}{e^{2t}+1}
$$
We take y(0)=y'(0)=0,so we can get:
$$
2c_1+2c_2+0.5\pi-ln2=0\\
2c_1+c_2+0.5\pi-0.5ln2=0
$$
So
$$
c_1=-0.25\pi,c_2=0.5ln2
$$
Therefore:
$$
y=-0.25\pi*e^{2t}+0.5ln2e^t+e^{2t}*arctan(e^t) - 0.5e^{t}*ln(e^{2t}+1)
$$

8
Term Test 2 / Re: Problem 4 (main sitting)
« on: November 19, 2019, 04:30:28 AM »
$$
det(A-{\lambda}I)=0\\

\begin{vmatrix}
3-\lambda & 3  \\
-2 & -1-\lambda
\end{vmatrix}={\lambda -1}^2+2=0
$$
WRONG So
$$
\lambda_1=1 +\sqrt{2}i\\
\lambda_2=1-\sqrt{2}i
$$
$$
\text{when } \lambda=1 +\sqrt{2}i\\
\begin{vmatrix}
2-\sqrt{2}i & 3  \\
-2 & -2-\sqrt{2}i
\end{vmatrix} = \begin{vmatrix}
0  \\
0
\end{vmatrix}
$$
RREF:
$$
\begin{pmatrix}
2-\sqrt{2}i & 3 \\
0 & 0
\end{pmatrix}
\quad = \begin{pmatrix}
0 \\
0
\end{pmatrix}
\quad
$$

Let x_2=t
So we can get:
$$
{(2-\sqrt{2}i})x_1=-3x_2=-3t\\
x_1=\frac{-3t}{2-\sqrt{2}i}
$$
So
$$
t*\begin{pmatrix}
-1-\frac{\sqrt{2}i}{2} \\
1
\end{pmatrix}
\quad
$$
Therefore:
$$
e^{1+i\sqrt2t}\begin{pmatrix}
-1-\frac{\sqrt{2}i}{2} \\
1
\end{pmatrix}
\quad =

e^t[\begin{pmatrix}
-cos(\sqrt{2}t)+\frac{\sqrt{2}}{2}*sin(\sqrt{2}t) \\
cos(\sqrt{2}t)
\end{pmatrix}
\quad +
i\begin{pmatrix}
-sin(\sqrt{2}t)-\frac{\sqrt{2}}{2}*cos(\sqrt{2}t) \\
sin(\sqrt{2}t)
\end{pmatrix}
\quad]
$$
So, general solution:
$$
y=c_1e^t\begin{pmatrix}
-cos(\sqrt{2}t)+\frac{\sqrt{2}}{2}*sin(\sqrt{2}t) \\
cos(\sqrt{2}t)
\end{pmatrix}
\quad +
c_2e^t\begin{pmatrix}
-sin(\sqrt{2}t)-\frac{\sqrt{2}}{2}*cos(\sqrt{2}t) \\
sin(\sqrt{2}t)
\end{pmatrix}
\quad
$$

9
Term Test 2 / Re: Problem 3 (main sitting)
« on: November 19, 2019, 04:29:59 AM »
(a):
We solve homo firstly:
$$
det(A-{\lambda}I)=0\\

\begin{vmatrix}
1-\lambda & 1  \\
-2 & 4-\lambda
\end{vmatrix}=-5\lambda+{\lambda}^2+6=0\\
(\lambda-2)(\lambda-3)=0\\
\lambda_1=2,\lambda_2=3
$$
Then:
$$
(A-{\lambda}I)x=0
$$
$$
When \lambda=2
$$
$$
\begin{pmatrix}
-1 & 1 \\
-2 & 2
\end{pmatrix}
\quad= \begin{pmatrix}
0  \\
0
\end{pmatrix}
\quad
$$
RREF:
$$
\begin{pmatrix}
-1 & 1 \\
 0 & 0
\end{pmatrix}
\quad= \begin{pmatrix}
0  \\
0
\end{pmatrix}
\quad
$$
$$
\text{Let x_2=t, so }x_1=x_2=t\\
t*\begin{pmatrix}
1 \\
1
\end{pmatrix}
\quad
$$
$$
When \lambda=3
$$
$$
\begin{pmatrix}
-2 & 1 \\
-2 & 1
\end{pmatrix}
\quad= \begin{pmatrix}
0  \\
0
\end{pmatrix}
\quad
$$
RREF:
$$
\begin{pmatrix}
 2 & -1 \\
 0 & 0
\end{pmatrix}
\quad= \begin{pmatrix}
0  \\
0
\end{pmatrix}
\quad
$$
$$
\text{Let x_2=t, so }x_1=0.5t, x_2=t\\
t*\begin{pmatrix}
1 \\
2
\end{pmatrix}
\quad
$$
So, the general solution is :
$$
y= c_1e^{2t}\begin{pmatrix}
1 \\
1
\end{pmatrix}
\quad +c_2e^{3t}\begin{pmatrix}
1 \\
2
\end{pmatrix}
\quad
$$




(b):
$$
\phi = \begin{pmatrix}
e^{2t} & e^{3t} \\
e^{2t} & 2e^{3t}
\end{pmatrix}
\quad
$$
$$
\phi * u' = g(t)
$$
$$
\begin{pmatrix}
e^{2t} & e^{3t} \\
e^{2t} & 2e^{3t}
\end{pmatrix}
\quad *{\begin{pmatrix}
u_1'  \\
u_2'
\end{pmatrix}
\quad}=\begin{pmatrix}
\frac{e^{4t}}{e^{2t} + 1 }  \\
0
\end{pmatrix}
\quad
$$
Simplify we can get:
$$
u_1'=-\frac{-e^t}{e^{2t} + 1}\\
u_2'=\frac{2e^{2t}}{e^{2t}+1}
$$
Therefore:
$$
u_1=ln(e^{2t}+1)+c_1\\
u_2=-arctane^t+c_2
$$
Finally:
$$
x=\phi * u=(ln(e^{2t}+1)+c_1)*{\begin{pmatrix}
e^{2t}  \\
e^{2t}
\end{pmatrix}
\quad} +(-arctane^t+c_2)*\begin{pmatrix}
e^{3t}  \\
2e^{3t}
\end{pmatrix}
\quad
$$
OK, except LaTeX sucks:

1) * IS NOT a sign of multiplication
2)  "operators" should be escaped: \cos, \sin, \tan, \ln

10
Term Test 2 / Re: Problem 2 (main sitting)
« on: November 19, 2019, 04:29:25 AM »
(a):
$$
W=ce^{-\int{p(t)}}dt\\
W=ce^{-\int{4}dt}\\
W=ce^{-4t}
$$



(b):
We can get:
$$
r^3+4r^2+r-6=0\\
(r-1)(r+2)(r+3)=0\\
r_1=1,r_2=-2,r_3=-3
$$
Therefore:
$$
y=c_1e^t+c_2e^{-2t}+c_3e^{-3t}
$$
$$
\begin{vmatrix}
e^t & e^{-2t} & e^{-3t} \\
e^t & -2e^{-2t} & -3e^{-3t} \\
e^t & 4e^{-2t} & 9e^{-3t}
\end{vmatrix}=-12e^{-4t}
$$
compare with (a) we get
$$
c=-12
$$


(c):
Let:
$$
y_p(t)=Ate^t\\
y'=A(e^t+te^t)\\
y''=A(2e^t+te^t)\\
y'''=A(3e^t+te^t)
$$
We take these into equation and get:
$$
A(3e^t+te^t)+4A(2e^t+te^t)+A(e^t+te^t)-6Ate^t=24e^t\\
12Ae^t=24e^t\\
A=2
$$
So:
$$
y_p(t)=2te^t
$$
Therefore:
$$
y=c_1e^t+c_2e^{-2t}+c_3e^{-3t}+2te^t
$$

OK. V.I.

11
Term Test 2 / Re: Problem 1 (main sitting)
« on: November 19, 2019, 04:29:03 AM »
(a):
$$
r^2+4=0\\
r_1=2i,r_2=-2i
$$
So we can get:
$$
y_c(t)=c_1cos2t+c_2sin2t
$$
$$
W=\begin{vmatrix}
cos2t & sin2t  \\
-2sin2t & 2cos2t
\end{vmatrix}=2
$$
$$
W_1=\begin{vmatrix}
0 & sin2t  \\
1 & 2cos2t
\end{vmatrix}=-sin2t
$$
$$
W_2=\begin{vmatrix}
cos2t & 0  \\
-2sin2t & 1
\end{vmatrix}=cos2t
$$
Therefore:
$$
Y(t)=cos2t\int{\frac{-sin2s*\frac{1}{(cos(s))^2}}{2}}ds+sin2t\int{\frac{cos2s*\frac{1}{(cos(s))^2}}{2}}ds\\
Y(t)=cos2t\int{-\frac{sins}{coss}}ds+sin2t\int{\frac{2(cos(s))^2-1}{2(cos(s)^2)}}ds\\
Y(t)=cost2t*lncost+sin2t*(t-0.5tant)
$$
So general solution is :
$$
y(t)=c_1cos2t+c_2sin2t+cost2t*lncost+sin2t*(t-0.5tant)
$$




(b):
$$
y'(t)=-2c_1sin2t+2c_2cos2t+cos2t*\frac{-sint}{cost} - 2sin2t*lncost+sin2t(1-0.5(sect)^2)+2cos2t(t-0.5tant)
$$
$$
\text{Take } y(0)=y'(0)=0\\
\text{We can get that: }c_1=c_2=0
$$
So finally:
$$
y=cost2t*lncost+sin2t*(t-0.5tant)
$$

OK, except LaTeX sucks:

1) * IS NOT a sign of multiplication
2)  "operators" should be escaped: \cos, \sin, \tan, \ln
$$
\boxed{  y=  \ln (\cos(t)) \cos(2t) + \Bigl(t-\frac{1}{2}\tan (t)\Bigr)\sin(2t). }
$$


12
Term Test 2 / Re: Problem 4 (morning)
« on: November 19, 2019, 04:28:39 AM »
$$
det(A-{\lambda}I)=0\\

\begin{vmatrix}
2-\lambda & -3  \\
4 & -2-\lambda
\end{vmatrix}={\lambda}^2+8=0
$$
So
$$
\lambda_1=\sqrt{8}i\\
\lambda_2=-\sqrt{8}i
$$
$$
\text{when } \lambda=\sqrt{8}I\\
\begin{vmatrix}
2-\sqrt{8}i & -3  \\
4 & -2-\sqrt{8}i
\end{vmatrix} = \begin{vmatrix}
0  \\
0
\end{vmatrix}
$$
RREF:
$$
\begin{pmatrix}
2-\sqrt{8}i & -3 \\
0 & 0
\end{pmatrix}
\quad = \begin{pmatrix}
0 \\
0
\end{pmatrix}
\quad
$$
Let x_2=t
So we can get:
$$
{(2-\sqrt{8}i})x_1=3x_2=3t\\
x_1=\frac{3t}{2-\sqrt{8}i}
$$
So
$$
t*\begin{pmatrix}
\frac{3}{2-\sqrt8i} \\
1
\end{pmatrix}
\quad
$$
Therefore:
$$
e^{i\sqrt8t}\begin{pmatrix}
\frac{3}{2-\sqrt8i} \\
1
\end{pmatrix}
\quad =
(cos(\sqrt8t)+isin(\sqrt8t))\begin{pmatrix}
\frac{3}{2-\sqrt8i} \\
1
\end{pmatrix}
\quad=\begin{pmatrix}
\frac{cos(\sqrt8t)-\sqrt2sin(\sqrt8t)}{2} +\frac{isin(\sqrt8t)+i\sqrt2cos(\sqrt8t)}{2}\\
cos(\sqrt8t )+ isin(\sqrt8t))
\end{pmatrix}
\quad
$$
So
$$
y=c_1\begin{pmatrix}
\frac{cos(\sqrt8t)-\sqrt2sin(\sqrt8t)}{2} \\
cos(\sqrt8t)
\end{pmatrix}
\quad +c_2\begin{pmatrix}
\frac{sin(\sqrt8t)+\sqrt2cos(\sqrt8t)}{2}\\
sin(\sqrt8t))
\end{pmatrix}
\quad
$$
OK, except LaTeX sucks:

1) * IS NOT a sign of multiplication
2)  "operators" should be escaped: \cos, \sin, \tan, \ln

13
Term Test 2 / Re: Problem 3 (morning)
« on: November 19, 2019, 04:28:19 AM »
(a):
$$
det(A-{\lambda}I)=0\\

\begin{vmatrix}
-2-\lambda & 1  \\
-1 & -\lambda
\end{vmatrix}=2\lambda+{\lambda}^2+1=0
$$
So
$$
(\lambda+1)^2=0\\
\lambda_1=\lambda_2=-1
$$
Then:
$$
(A-{\lambda}I)x=0
$$
$$
\begin{pmatrix}
-1 & 1 \\
-1 & 1
\end{pmatrix}
\quad= \begin{pmatrix}
0  \\
0
\end{pmatrix}
\quad
$$
RREF
$$
\begin{pmatrix}
-1 & 1 \\
0 & 0
\end{pmatrix}
\quad= \begin{pmatrix}
0  \\
0
\end{pmatrix}
\quad
$$
$$
\text{if let }x_2=t\\
x_1=x_2=t
$$
So
$$
t*{\begin{pmatrix}
1  \\
1
\end{pmatrix}
\quad}
$$
Since we just have only one eigenvector:
$$
\begin{pmatrix}
-1 & 1 \\
-1 & 1
\end{pmatrix}
\quad= \begin{pmatrix}
1  \\
1
\end{pmatrix}
\quad
$$
$$
\begin{pmatrix}
-1 & 1 \\
0 & 0
\end{pmatrix}
\quad= \begin{pmatrix}
1  \\
0
\end{pmatrix}
\quad
$$
$$
\text{So the other eigenvector is:}= \begin{pmatrix}
0  \\
1
\end{pmatrix}
\quad
$$
Therefore:
$$
y=c_1e^{-t}*{\begin{pmatrix}
1  \\
1
\end{pmatrix}
\quad}+c_2e^{-t}[t*{\begin{pmatrix}
1  \\
1
\end{pmatrix}
\quad} +{\begin{pmatrix}
0  \\
1
\end{pmatrix}
\quad}]
$$




(b):
$$
\phi = \begin{pmatrix}
e^{-t} & te^{-t} \\
e^{-t} & e^{-t}(t+1)
\end{pmatrix}
\quad
$$
$$
\phi * u' = g(t)
$$
$$
\begin{pmatrix}
e^{-t} & te^{-t} \\
e^{-t} & e^{-t}(t+1)
\end{pmatrix}
\quad *{\begin{pmatrix}
u_1'  \\
u_2'
\end{pmatrix}
\quad}=\begin{pmatrix}
0  \\
\frac{e^{-t}}{t^2 + 1 }
\end{pmatrix}
\quad
$$
Simplify we can get:
$$
u_1'=-\frac{t}{t^2 + 1}\\
u_2'=\frac{1}{t^2+1}
$$
Therefore:
$$
u_1=-0.5ln(t^2+1)+c_1\\
u_2=arctant+c_2
$$
Finally:
$$
x=\phi * u=(-0.5ln(t^2+1)+c_1)*{\begin{pmatrix}
e^{-t}  \\
e^{-t}
\end{pmatrix}
\quad} +(arctant+c_2)*\begin{pmatrix}
te^{-t}  \\
e^{-t}(t+1)
\end{pmatrix}
\quad
$$

14
Term Test 2 / Re: Problem 2 (morning)
« on: November 19, 2019, 04:27:56 AM »
No double-dipping
(a):
$$
W=ce^{-\int{p(t)}}dt\\
W=ce^{-\int{-2}dt}\\
W=ce^{2t}
$$



(b):
We can get:
$$
r^3-2r^2+4r-8=0\\
(r-2)(r^2+4)=0\\
r_1=2,r_2=2i,r_3=-2i
$$
Therefore:
$$
y=c_1e^2t+c_2sin2t+c_3cos2t
$$
$$
\begin{vmatrix}
e^2t & sin2t & cos2t \\
2e^2t & 2cos2t & -2sin2t \\
4e^2t & -4sin2t & -4cos2t
\end{vmatrix}=e^{2t}({-8(cos2t)^2}-8(sin2t)^2)-sin2t(-8e^{2t}cos2t+8e^{2t}sin2t)+cos2t(-8e^{2t}sin2t-8e^{2t}cos2t)=-8e^{2t}-8e^{2t}=-16e^{2t}
$$
compare with (a) we get
$$
c=-16
$$



(c):
Let
$$
y=Acost+Bsint
$$
Therefore:
$$
y'=-Asint+Bcost\\
y''=-Acost-Bsint\\
y'''=Asint-Bcost
$$
Next We take these into equation and get:
$$
(-3A-6B)sint+(3B-6A)cost=15cost\\
-3A-6B=0:3B-6A=15\\
A=-2,B=1
$$
Finally:
$$
y=c_1e^2t+c_2sin2t+c_3cos2t-2cost+sint
$$

15
Term Test 2 / Re: Problem 1 (morning)
« on: November 19, 2019, 04:27:31 AM »
You should not double-dip. V.I.
The first question
we solve homo firstly, so we get:
$$
r^2 - 1=0\\
r^2 =1\\
r_1=1\\
r_2=-1\\
\text{Therefore: }y=c_1e^t+c_2e^{-t}
$$
So we can get
$$
W = \begin{vmatrix}
e^t & e^{-t} \\
e^t & -e^{-t}
\end{vmatrix}=-2\\
W_1=\begin{vmatrix}
0 & e^{-t} \\
1 & -e^{-t}
\end{vmatrix}=-e^{-t}\\
W_2=\begin{vmatrix}
e^t & 0 \\
e^t& 1
\end{vmatrix}=e^t
$$
So we can get
$$
Y(t)=e^t\int{\frac{-e^{-s}*\frac{12}{e^s+1}}{-2}}ds + e^{-t}\int{\frac{e^{s}*\frac{12}{e^s+1}}{-2}}ds\\
Y(t)=6e^t\int{\frac{e^{-s}}{e^s+1}}ds - 6e^{-t}\int{\frac{e^{s}}{e^s+1}}ds\\
Y(t)=6e^t\int{\frac{1}{e^s*(e^s+1)}}ds - 6e^{-t}\int{\frac{e^{s}}{e^s+1}}ds\\
Y(t)=6e^t\int{\frac{e^s+1-e^s}{e^s*(e^s+1)}}ds - 6e^{-t}\int{\frac{e^{s}}{e^s+1}}ds\\
Y(t)=6e^t\int{\frac{1}{e^s}-\frac{1}{e^s+1}}ds - 6e^{-t}\int{\frac{e^{s}}{e^s+1}}ds\\
Y(t)=6e^t\int{\frac{1}{e^s}-\frac{e^s+1-e^s}{e^s+1}}ds - 6e^{-t}\int{\frac{e^{s}}{e^s+1}}ds\\
$$
Finally:
$$
Y(t)=6e^t[-e^{-t}-t+ln(e^t+1)] - 6e^{-t}ln(e^t+1)\\
y(t)=c_1e^t+c_2e^{-t}+6e^t[-e^{-t}-t+ln(e^t+1)] - 6e^{-t}ln(e^t+1)
$$


The second question:
since we have y(t), so we can get y'(t)
$$
y'(t)=c_1e^t-c_2e^{-t}+6e^t[-e^{-t}-t+ln(e^t+1)]+6e^t[e^{-t}-1+\frac{e^t}{e^t+1}]+6e^{-t}ln(e^t+1)-6e^{-t}*\frac{e*t}{e^t+1}
$$
Then we take y(0)=y'(0)=0 into y and y'
Therefore:
$$
c_1+c_2-6=0\\
c_1-c_2-6+12ln2=0\\
c_1=6-6ln2, c_2=6ln2
$$
So
$$
y(t)=(6-6ln2)e^t+6ln2e^{-t}6e^t[-e^{-t}-t+ln(e^t+1)] - 6e^{-t}ln(e^t+1)
$$

Pages: [1] 2