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Messages - Rudolf-Harri Oberg

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Quiz 4 / Re: Quiz 4--Problem (night sections)
« on: March 20, 2013, 09:04:17 PM »
a)For critical points, we just set $x'=0$ and $y'=0$ which will yield four critical points: $P1=(0,0), P2=(-1,-1), P3=(2,2), P4=(-2,0)$. For linearization, we have to compute the Jacobian i.e the matrix of first derivatives of the functions. If $F=(2+x)(y-x), G=y(2+x-x^2)$, then $F_x=y-2x-2, F_y=2+x, G_x=y(1-2x), G_y=2+x-x^2$. Now we have got everything to find corresponding linearized systems, just use equation 13 from page 522 (book).

b) To determine the nature of solutions at each critical point, we just have to find the eigenvalues of the matrices we get from evaluating the Jacobian at the respective critical points.

For the point $P1$, we evaluate the Jacobian to find that the resulting matrix has eigenvalues $r_1=2, r_2=-2$, so the the critical point is a saddle (look picture from handout), this is unstable.

For the point $P2$, we evaluate the Jacobian to find that the resulting matrix has complex eigenvalues (the characteristic equation is $r^2+r+3=0$), so the the critical point is a spiral point, this is (asymptotically) stable as the real part of the eigenvalues is negative.

For the point $P3$, we evaluate the Jacobian to find that the resulting matrix has complex eigenvalues (the characteristic equation is $r^2+4r+24=0$), so the the critical point is a spiral point, this is (asymptotically) stable as the real part of the eigenvalues is negative.

For the point $P4$, we evaluate the Jacobian to find that the resulting matrix has eigenvalues $r_1=2, r_2=-4$, so the the critical point is a saddle, this is unstable.

For visualization of solutions, go to the course homepage and look at the section of learning resources. Pick the link to math.rice, select PPLANE.2005.10, type in the equations and enjoy the picture! You can also see the answer to the bonus question from the picture.

MidTerm / Re: MT Problem 3
« on: March 06, 2013, 10:50:34 PM »
This is an Euler equation, see book page 166, problem 34. We need to use substitution $x=\ln t$, this will make into a ODE with constant coefficients. We look first at the homogenous version:

Solving $r^2-3r+2=0$ yields $r_1=2, r_2=1$. So, solutions to the homogeneous version are $y_1(x)=e^{2x}, y_2(x)=e^{x}$. But then solutions to the homogeneous of the original problem are $y_1(t)=t^2, y_2(t)=t$.
So, $Y_{gen.hom}=c_1t^2+c_2t$. We now use method of variation of parameters, i.e let $c_1,c_2$ be functions.
To use the formulas on page 189, we need to divide the whole equation by $t^2$ so that the leading coefficient would be one, so now $g=t e^t$. The formula is:
$c_i'=\frac{W_i g}{W}$, where $W_i$ is the wronksian of the two solutions where the i-th column has been replaced by $(0,1)$.
We now just calculate that $W(t^2,t)=-t^2, W_1=-t, W_2=t^2$. Now we need to compute $c_1, c_2$.

$$c_1'=e^t \implies c_1=e^t$$
$$c_2'=-te^t \implies c_2=-e^t(t-1)$$

Plugging these expressions back to $Y_{gen.hom}$ yields the solution which is

MidTerm / Re: MT Problem 4
« on: March 06, 2013, 10:12:21 PM »
We use Milman's method:

$$L\left[A\frac{x^m}{m!}e^{rx}\right]=Ae^{rx}\left(Q(r)\frac{x^m}{m!}+Q'(r)\frac{x^{m-1}}{(m-1)!}+Q''(r)\frac{x^{m-2}}{2!(m-2)!}+...  \right)$$
In this case $Q=r^3-2r^2+4r-8$. As we see an exponent in the power of three, let $r=3$. We need to evaluate $Q(3)=13$. As there are no polynomial terms, let $m=0$.

Then $L(Ae^{3x})=13Ae^{3x}$ which implies $A=\frac{1}{13}$.

Solution is $Y_p=\frac{1}{13}e^{3x}$.

MidTerm / Re: MT Problem 5
« on: March 06, 2013, 10:03:41 PM »
We begin by finding eigenvalues for the systems matrix. We solve $(5-\lambda)(-4-\lambda)+18=\lambda^2-\lambda-2=0$. This yields $\lambda_1=2, \lambda_2=-1$. We now search for eigenvectors.

For $\lambda_1=2$, the eigenvector is $\xi_1=(1,1)$

For $\lambda_2=-1$, the eigenvector is $\xi_2=(1,2)$.

General solution for the system is $Y_G=c_1e^{2t}\xi_1+c_2e^{-t}\xi_2$

MidTerm / Re: MT Problem 1
« on: March 06, 2013, 09:57:28 PM »
We use Milman's method to find a particular solution:
$$L\left[A\frac{x^m}{m!}e^{rx}\right]=Ae^{rx}\left(Q(r)\frac{x^m}{m!}+Q'(r)\frac{x^{m-1}}{(m-1)!}+Q''(r)\frac{x^{m-2}}{2!(m-2)!}+...  \right)$$
In this case
$Q=r^2-3r+2, Q'=2r-3, Q''=2$. We want to have $r=3$. So we evaluate that $Q(3)=2, Q'(3)=3, Q''(3)=2$.
Also, let $m=0$. Then $L(Ae^{3x})=2Ae^{3x}$. This implies that $A=1$.
In conclusion, solution is $Y_p=e^{3x}$.

For general solution to homogeneous equation, solve $r^2-3r+2=0$ yielding $r_1=2$, $r_2=1$.
So, $Y_{gen.hom}=c_1e^{2x}+c_2e^{x}$

General solution for the whole system is $Y_G=Y_{gen.hom}+Y_p=c_1e^{2x}+c_2e^{x}+e^{3x}$.
Condition $Z(0)=1$ yields that $c_1+c_2+1=1$. Condition $Z'(0)=0$ yields that $2c_1+c_2+3=0$. Solving these gives $c_1=-3, c_2=3$.

In conclusion we get that: $y=-3e^{2x}+3e^x+e^{3x}$.

Quiz 3 / Re: Night Sections Problem 2
« on: February 27, 2013, 10:17:22 PM »
We start by solving $r^3-r=0$ which gives that $r_1=0, r_2=1, r_3=-1$.
Variation of parameters is not a good method to guess a particular solution here. You can try guessing that the particular solution is $Y_p=A\sin(t)+B\cos(t)$ or just look at the equation and deduce that $Y_p=\cos(t)$

So, general solution to the equation is

Quiz 3 / Re: Night Sections Problem 1
« on: February 27, 2013, 10:13:15 PM »
We first solve the characteristic polynomial: $r^3+r=0$ which yields $r_1=0, r_{2,3}=+- i$. So, solution to the general homogeneous equation is
$$y_g=u_1+u_2\cos(t)+u_3 \sin(t)$$
Idea in method of parameters is to let $u_1,u_2,u_3$ to be functions and then plot the $y_g$ into the equation (in this case compute it's first and third derivative), solving for $u_1,u_2,u_3$.
This can be done, but textbook page 242 gives you a nice final formula: $$u'_m=\frac{gW_m}{W}$$

So, we use this - maybe this was the reason why I finished 7 minutes earlier.

It turns out that $W(1,\cos(t), \sin(t))=1$. $W_m$ is just the same wronskian but the m-th column is replaced by the column $(0,0,1)$. With this definition
$$W_1=1, W_2=-\cos(t), W_3=-\sin(t)$$
Using the previously stated formula and the fact that in this case $g=\tan(t)$:

$$u_1'=\tan(t) \implies u_1=-\ln(\cos(t))$$
$$u_2'=-\tan(t)\cos(t)=-\sin(t)\implies u_2=\cos(t)$$
$$u_3'=-\tan(t)\sin(t)=-\left(\frac{1-\cos^2(t)}{\cos(t)} \right)=-\left(\frac{1}{\cos(t)}-\cos(t)\right)\implies
The third integral was the hardest. But in the homework problems you also had to compute $\int \frac{dt}{\cos(t)}$ which is surprisingly difficult. So, I had worked on the integral before and remembered the solution.

Now, we just have to plug $u_1,u_2,u_3$ back to the general solution of homogeneous equation $y_g$ and we obtain that the particular solution is:
General solution is therefore
$$Y_G=Y_p+c_1+c_2\cos(t)+c_3 \sin(t)$$

MAT 244 Misc / Re: Advice on how to get faster?
« on: February 14, 2013, 12:44:14 PM »
I solved nearly every homework problem, I knew the theory well. Yet, I ran into computational difficulties and were not able to fully finish questions (i.e differentiate enough times to solve for constants, evaluate integrals etc)

From the given feedback so far and according to my personal experience, the message is clear: the questions selected for the test were not student-friendly or even unsuitable for a test of such a format (limited space, tight time constraints).

I feel somewhat deceived by the staff, instructors. Someone had not prepared for that Wednesday night and in my view it was not the students.

I hope the next test goes better.

Term Test 1 / Re: TT1--Problem 2
« on: February 14, 2013, 12:08:10 PM »
Part B, Alternative solution.
Assume the second solution is of the form $y_2=z(t)t$. Plugging that into the ODE, we get that
$tz''+(2+t \frac{-t \cos (t)}{\cos(t)+t \sin(t)})z'=0$ (look at reduction of order at pg 171, formula 30). This is a first order ODE for z'. So, plug for example $u=z'$. This equation is separable and simplifies to

$$tu'=(\frac{t^2 \cos(t)}{\cos(t)+t \sin(t)}-2)u$$
$$\frac{1}{u}du=\frac{1}{t}(\frac{t^2 \cos(t)}{\cos(t)+t \sin(t)}-2)dt$$.
Now "just" integrate and you get
$$\ln u= \ln(t\sin(t)+\cos(t))-2\ln (t) + \ln C_1$$ or
$$z'=u=\frac{t\sin(t)+\cos(t)}{t^2}$$ (forget the constant)
Now we have to integrate once again to get
$z=\frac{-\cos(t)}{t}+C$; so in conclusion we get that $y_2=zy_1=\frac{-\cos(t)}{t}t+Ct=-cos(t)+Ct$

We need $y_2(\frac{\pi}{2})=0$ which yields C=0. So, $y_2=-\cos(t)$ is the function we are looking for. Also note that 
$W(y_1,y_2)(\frac{\pi}{2})=-\cos(\frac{\pi}{2})+\frac{\pi}{2}\sin (\frac{\pi}{2})=\frac{\pi}{2}$, so the Wronskian condition is also satisfied.

Comment: I went for this solution idea during my test. Unfortunately, given half a page of room, exam pressure and those nasty integrals, I could not get to the final solution.

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