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Messages - Tim Mengzhe Geng

Pages: [1] 2
1
Final Exam / Re: FE-P4
« on: April 14, 2018, 03:01:14 AM »
The integration of 1/(cos t sint) is ln(sint) - ln(cost) + c
and hence you get ln(tant) + c
I'm so sorry that you're also correct  and the two result are the same since
$\tan\frac{x}{2}=\frac{\sin(x)}{1+cos(x)}$
and therefore
$-\ln|csc(2x)+cot(2x)|=\ln|tan(x)|$

2
Final Exam / Re: FE-P4
« on: April 14, 2018, 02:42:24 AM »
The integration of 1/(cos t sint) is ln(sint) - ln(cost) + c
and hence you get ln(tant) + c
$1/(costsint)=2/sin(2t)=2csc(2t)$
And by the integration of 2csc(2t),
we get
$-\ln|csc(2t)+cot(2t)|+c$

3
Final Exam / Re: FE-P1
« on: April 14, 2018, 02:40:02 AM »
There is a  mistake in step (9)
the first term should be 2y(x^2)cosy instead of 2x(y^2)cosy
Thanks and I've changed it

4
Final Exam / Re: FE-P3
« on: April 12, 2018, 10:01:46 PM »
Since the solution is incomplete after Y(x),
I am attaching a copy of my solution
Sorry what do you mean by "the solution is incomplete after Y(x)"
I did write a bit more on the exam (expanding the summation) but I think one should be able to get full marks if he integrates everything and mention how the solution is composed.(Given that the integral is correct)

5
Final Exam / Re: FE-P4
« on: April 12, 2018, 03:32:57 PM »
since the above solution is incomplete after μ(2)
I am attaching a copy of my solution...
I think your integration to $\mu_2(t)$may not be correct

6
Final Exam / Re: FE-P1
« on: April 12, 2018, 09:56:15 AM »
I think the solution is wrong, my h(y) was coming out to be just a constant, c
I checked my solution again and didn't find something wrong. So maybe we should leave it to the Prof.

7
Final Exam / Re: FE-P6
« on: April 12, 2018, 09:51:12 AM »
For Part(a), Note that for stationary points, we should have

x^2+y^2-1=0

And at the same time

-2xy=0

Therefore there're totally four stationary points. They are

(x,y)=(1,0), (-1,0), (0,1) or (0,-1).

J={
\left[\begin{array}{ccc}
2x & 2y \\
-2y & -2x
\end{array}
\right ]},

At point (1,0),

J[1,0]={
\left[\begin{array}{ccc}
2 & 0 \\
0 & -2
\end{array}
\right ]},

At point (-1,0),

J[-1,0]={
\left[\begin{array}{ccc}
-2 & 0 \\
0 & 2
\end{array}
\right ]},

At point (0,1),

J[0,1]={
\left[\begin{array}{ccc}
0 & 2 \\
-2 & 0
\end{array}
\right ]},

At point (0,-1),

J[0,-1]={
\left[\begin{array}{ccc}
0 & -2 \\
2 & 0
\end{array}
\right ]},

8
Final Exam / Re: FE-P6
« on: April 12, 2018, 09:22:44 AM »
there is a small mistake..... in step 5 you have mentioned that h(y) = 0
it is notzero, it is a constant
Please note that at I state, I just choose $h(y)=0$ for simplification.

9
Final Exam / Re: FE-P6
« on: April 12, 2018, 12:41:28 AM »
For part(b), we have

(x^2+y^2-1)dx+2xydy=0

Note that

M_y=N_x=2y

The equation is exact.
By integration

H=\frac{1}{3}x^3+xy^2-x+h^\prime(y)

h^\prime(y)=0

We choose

h(y)=0

In this way,

H(x,y)=\frac{1}{3}x^3+xy^2-x=C

I will post solution to other parts later if no one else follows.

10
Final Exam / Re: FE-P5
« on: April 12, 2018, 12:34:35 AM »
For part(a)
Let

x(x-y+1)=0

and

y(x-2)=0

We will have all the three critical points

(x,y)=(0,0),(2,3) or (-1,0)

For part(b)

F=x(x-y+1)

G=y(x-2)

Therefore, the Matrix $J$

J={
\left[\begin{array}{ccc}
2x-y+1 & -x \\
y & x-2
\end{array}
\right ]},

At point(0,0), we have

J[0,0]={
\left[\begin{array}{ccc}
1 & 0 \\
0 & -2
\end{array}
\right ]},

Eigenvalues are

\lambda_1=-2

\lambda_2=1

Therefore, (0,0) is a saddle point and thus unstable.
At point(2,3), we have

J[2,3]={
\left[\begin{array}{ccc}
2 & -2 \\
3 & 0
\end{array}
\right ]},

\lambda_3=1+\sqrt{5}i

\lambda_4=1-\sqrt{5}i

Therefore, (2,3) is a spiral point and is unstable. The orientation is counterclockwise.

J[-1,0]={
\left[\begin{array}{ccc}
-1 & 1 \\
0 & -3
\end{array}
\right ]},

\lambda_5=-1

\lambda_6=-3

Therefore, (-1,0) is a node and is asymptotically stable.

11
Final Exam / Re: FE-P4
« on: April 12, 2018, 12:16:56 AM »

A={
\left[\begin{array}{ccc}
1 & 1 \\
-1 & 1
\end{array}
\right ]},

The eigenvalues are

\lambda_1=1+i

\lambda_2=1-i

The eigenvector corresponding to $\lambda_1$ is

\xi^(1)=(1,i)^{T}

Then we can have the solution to the homogeneous system

(x,y)^T=c_1\cdot (e^t\cos(t), -e^t\sin(t))^T+c_2\cdot(e^t\sin(t), e^t\cos(t))^T

We can get a fundamental matrix from here

\Psi={
\left[\begin{array}{ccc}
e^t\cos(t) & -e^t\sin(t) \\
e^t\sin(t) & e^t\cos(t)
\end{array}
\right ]},

Note that we have

g(t)=(\frac{e^t}{\cos(t)},\frac{e^t}{\sin(t)})^T

And suppose

\Psi \mu^\prime=g(t)

We shoud have for the nonhomogeneous system

(x,y)^T=\Psi \mu

This is the method of Variation of Parameters. By plugging in and integration, we have

\mu_1(t)=c_3

\mu_2(t)=-\ln|cot(2t)+csc(2t)|+c_4

Finally we get the required solution according to (6), (9), (10) and (11)

12
Final Exam / Re: FE-P3
« on: April 11, 2018, 11:57:57 PM »
Also for $(20)$, $\ln(e^x)$ can be simplified as $\ln(e^x)=x$.
Thanks again and I will modify it

13
Final Exam / Re: FE-P3
« on: April 11, 2018, 11:49:31 PM »
Small Error: $W_2(x)$ should be $2e^{4x}$.
For this case I don't think so since when we expand the matrix, we have to times $(-1)^{i+j}$

14
Final Exam / Re: FE-P2
« on: April 11, 2018, 11:38:20 PM »
Most of the answer is correct, there is one small computational error.$\\$
For $Y_2(t)$, it should be $Y_2(t)=-4\cos(t)+2\sin(t)$.$\\$
Thanks and I will modify it

15
Final Exam / Re: FE-P3
« on: April 11, 2018, 11:37:34 PM »
First we find the solution for the homogeneous system

y^{(3)}-6y^{(2)}+11y^{(1)}-6y=0

The corresponding characteristic equation is

r^3-6r^2+11r-6=0

Three roots are

r_1=1

r_2=2

r_3=3

Then the solution for the homogeneous system is

y_c(t)=c_1e^{x}+c_2e^{2x}+c_3e^{3x}

where

y_1(t)=e^{x}

y_2(t)=e^{2x}

y_3(t)=e^{3x}

Then we follow to find the required solution to the nonhomogeneous equation. We use Variation of Parameters. We have

W[y_1,y_2,y_3]=2e^{6x}

W_1[y_1,y_2y_3]=e^{5x}

W_2[y_1,y_2y_3]=-2e^{4x}

W_3[y_1,y_2，y_3]=e^{3x}

and

g(x)=2\frac{e^{3x}}{e^{x}+1}

And then we have the following integration

\int \frac{W_1\cdot g(x)dx}{W[y_1,y_2,y_3]}=\int\frac{e^{2x}}{e^{x}+1}

\int\frac{e^{2x}}{e^{x}+1}=e^{x}-\ln(e^{x}+1)+c_4

\int \frac{W_2\cdot g(x)dx}{W[y_1,y_2,y_3]}=-2\int\frac{e^{x}}{e^{x}+1}

-2\int\frac{e^{x}}{e^{x}+1}=-2\ln(e^{x}+1)+c_5

\int \frac{W_3 \cdot g(x)dx}{W[y_1,y_2,y_3]}=\int\frac{1}{e^{x}+1}

\int\frac{1}{e^{x}+1}=x-\ln(e^{x}+1)+c_6

And finally, the required general solution $y(t)$

y(t)=\sum_{i=1}^3 y_i(t)\cdot\int\frac {W_i\cdot g(x)dx}{W[y_1,y_2,y_3]}

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