Recent content by Zem

The linear algebra explanation of this is the easiest way for me to understand it. Since (A - \lambda I = 0) is -2a - 0b = 0, a is a free variable. So a could be anything, which leads to an eigenvector that could be anything. Thanks everyone!

After asking the professor about this, it turns out the matrix is \left(\begin{array}{c} 0 \ 0\\ 0 \ 0 \end{array}\right) after (A-\lambda I){\bf v}=0 This means any vector will work. v_1 = \left(\begin{array}{c} 1 \\ 0 \end{array}\right) v_2 = \left(\begin{array}{c} 0 \\ 1...

(A-\lambda I){\bf v}=0 \left(\begin{array}{c} -2 \ 0\\ 0 \ -2 \end{array}\right) \left(\begin{array}{c} -2-(-2) \ 0\\ 0 \ -2-(-2) \end{array}\right) \left(\begin{array}{c} 0 \ 0\\ 0 \ 0 \end{array}\right) or (-2 - (-2)){\bf v}=0 0v = 0 What am I missing?

Oops. \lambda ^2 + 4\lambda + 4 = 0 \lambda_1 = \lambda_2 = -2 Find v_1: (A - \lambda I = 0) \left(\begin{array}{c} 0 \ 0\\ 0 \ 0 \end{array}\right) How can I get an eigenvector when (A - lambda) = 0? Since the eigenvalues are equal and less than zero, I know it is a stable proper...

I got the eigenvalues with the characteristic equation. (-2 - \lambda)(-2 - \lambda) -4 = 0 \lambda^2 + 4\lambda = 0 \lambda_1 = 0, \lambda_2 = -4 Case 1: (A - \lambda I = 0) -2x + 0y = 0 0x - 2y = 0 x = 0, y =0 v_1 = [0,0] Case 2: 2x + 0y = 0 0x + 2y = 0 v_2 = [0,0] Both...

I am getting stuck on the very beginning of these homework problems. Solve the linear system to determine whether the critical point (0,0) is stable, asymtotically stable, or unstable. dx/dt = -2x, dy/dt = -2y The book uses separation of variables, but the professor has instructed us to use...

I'd like to get the text for the class from the library, but it's checked out. Nr.com is awesome, exactly what I think I need. Thanks!

I am a geology major interested in groundwater modeling. Currently taking ordinary differential equations, and I'll be in a computational numerical methods course for engineers next fall. Here is the course description.. "Introduction to numerical methods for environmental engineering...

Thanks! I have found the radius of convergence and general solution of the second problem. Back to the first problem. Note: (x^2 + 1)y'' + 6xy' + 4y = 0 (\sum_{n = 2}^{\infty} \{n(n-1)C_n*X^n}) + (\sum_{n = 2}^{\infty} \{n(n-1)C_n*X^{{n-2}}) + 6(\sum_{n=1}^{\infty} \{n}{C_n}*X^n}) +...

Yes, so I'll make my patterns... C_{n+2}=\frac{-C_{n}}{(n+2)} n=0: C_{2} = \frac{-C_{0}}{2} n=1: C_{3} = \frac{-C_{1}}{3} n=2: C_{4} = \frac{(-1)C_{0}}{2*4} n=3: C_{5} = \frac{(-1)C_{1}}{3*5} n=4: C_{6} = \frac{(-1)C_{0}}{4*6} The even ones look like C_{2n} = \frac{(-1)^nC_{0}}{2^{n}n!}...

These are differential equations problems. They are now "redo's", so I have hints from the grader that I don't understand. First problem: (x^2 + 1)y'' + 6xy' + 4y = 0 After isolating C_n+2, I have this. (n+2)(n+1)C_n+2 * X^n = -n(n-1)C_n * X^n - 6nC_n * X^n + 4C_n C_n+2 = [-n(n-1)C_n - 6nC_n +...

Yes, I was just stuck in the middle of the problem. Now all I have to do is put the real and complex parts of the solutions together to get the general solutions of x_1 and x_2. Piece of cake. I am curious, though, what you were saying about T and F. Is that stuff for linear algebra? We did...

My problem was that I was stuck on assigning b = 1, but it appears much better to make b = a's coefficient. When I set b = (2-4i), a = 5, so V_1 = \left(\begin{array}{cc}5\\2-4i\end{array}\right) :cool:

Find a general solution of the given system using the method (A - \lambdaI)V2 = V1. x'_1 = 2x_1 - 5x_2, x'_2 = 4x_1 - 2x_2 x' = \left(\begin{array}{cc}2&-5\\4&-2\end{array}\right) characteristic equation: (2 - \lambda)((-2) - \lambda) + 20 = 0 \lambda^2 + 16 = 0 \lambda = 4i Using this...

Thank You!