Recent content by kleyton

This is exactly what I was intending. Your geometric method does seem interesting, but I do not quite seem to understand it. The above is what you had tried to show in your post for 3^2. Could you explain what you meant by "delta". The image seem to show 1+3+5=9. That is what the...

I have been working on representing the powers of numbers as a summation. This is as far as I have gotten. Power: 2 m^2 = \sum_{n=1}^m \left(2n -1\right) Power: 3 m^3 = \sum_{n=1}^m \left(3n^2 -3n +1\right) Power: 4 m^4 = \sum_{n=2}^m \left[6*(4n-6) * \left(\sum_{a=1}^{m-n+1}...

@HallsofIvy I cannot believe I did not see that. I learned that in differential equations class and even used it often in adv engg math. Thanks a ton. This just makes solving the problem a whole lot easier (might I say, cleaner too). Kudos to you ! I will post the completed solution once I...

Thanks a ton micromass ! Your pointers were very helpful. Just one last question. If the equations were not homogeneous, which would be the simplest way to solve it, undetermined coefficients or variation of parameters ? PS: I tried to follow the example on matrix differential equation on...

I am not sure what you meant by 'diagonizable'. But, I tried solving for the eigenvectors as follows (wiki): \begin{pmatrix} 1 & -1 \\ -4 & 1 \end{pmatrix}\begin{pmatrix} \alpha \\ \beta \end{pmatrix} = \lambda \begin{pmatrix} \alpha \\ \beta \end{pmatrix} This gave me, \hat{v_{1}} =...

Thanks for pointing that out. Never looked at the problem that way. I think it is a eigenvalue problem. \begin{vmatrix}\begin{bmatrix} 1 & -1 \\ -4 & 1 \end{bmatrix}- \lambda \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\end{vmatrix} = 0 And I solve for lambda and get \lambda_{1} = 3...

Homework Statement Solv the simultaneous homogenous differential equation \begin{cases} dy/dx + dz/dx + 3y = 0 \\ dy/dx - y + z = 0 \end{cases} Homework Equations The Attempt at a Solution (from eq 2), dy/dx = y - z --- eq (3) (substituting eq 3 in eq 1), \therefore dz/dx = z...