Toronto Math Forum

$$
\int_{\Upsilon}e^z dz
$$
$$
\text{line from 0 to } z_{0}$$
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
$$
\int_{\Upsilon}e^z dz =  \int_{0}^{1}f(r(t))r'(t)dt = \int_{0}^{1}e^{tz_{0}}z_{0} dt = z_{0}(\frac{1}{z_{0}}e^{z_{0}}- \frac{1}{z_{0}})= e^{z_{0}} - 1
$$

Sorry sir, , could you point it out?

I have modified it sir.

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:
$$
y=-0.25\pi*e^{2t}+0.5ln2e^t+e^{2t}*arctan(e^t) - 0.5e^{t}*ln(e^{2t}+1)
$$

(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

(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
$$
\boxed{  y=  \ln (\cos(t)) \cos(2t) + \Bigl(t-\frac{1}{2}\tan (t)\Bigr)\sin(2t). }
$$

(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:
$$
x=\phi * u=(-0.5ln(t^2+1)+c_1)*{\begin{pmatrix}
e^{-t}  \\
e^{-t}
\end{pmatrix}
\quad} +(arctant+c_2)*\begin{pmatrix}
te^{-t}  \\
e^{-t}(t+1)
\end{pmatrix}
\quad
$$

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
$$
y(t)=(6-6ln2)e^t+6ln2e^{-t}6e^t[-e^{-t}-t+ln(e^t+1)] - 6e^{-t}ln(e^t+1)
$$