Recent content by CentreShifter

Excellent. Thank you.

This is surely the simplest problem imaginable in DE, but it's been a few years and I'm having trouble recalling. The goal of my task doesn't necessitate relearning DE, so I thought I would take a shot at asking directly. Simply, I wish to express the time-dependent rate equation...

I've uploaded my attempted solution here. The problem I'm having is highlighted at the bottom in red. The issue I'm having is expressing the direction of H. I realize the cancellation that occurs at point (0,0,z), where the only the z-component of the H-field remains. I also realize that my...

This is just a general question regarding Boolean minimization. Expression: Y=\bar{X}_1\bar{X}_0+\bar{X}_1X_0+X_1\bar{X}_0 Minimized expression: Y=\bar{X}_1+\bar{X}_0 My first attempt was to minimize it algebraically. I factored \bar{X}_1 from the first two terms, then the \bar{X}_0+X_0...

Problem: Show, without evaluating directly, that \left|\begin{matrix} a_1+b_1t&a_2+b_2t&a_3+b_3t \\ a_1t+b_1&a_2t+b_2&a_3t+b_3 \\ c_1&c_2&c_3 \end{matrix}\right| = (1-t^2)\left|\begin{matrix} a_1&a_2&a_3 \\ b_1&b_2&b_3 \\ c_1&c_2&c_3 \end{matrix}\right| Clearly, here I'm...

Beautiful. Inverting both sides did the trick. Inverse of the RHS is \begin{bmatrix}\frac{-5}{13} & \frac{2}{13} \\ \frac{4}{13} & \frac{1}{13} \end{bmatrix}. I'll call this B^{-1} So then I'm left with \begin{align}I+2A&=B^{-1} \\ 2A&=B^{-1}-I \\...

Inverse of a sum of matrices [solved] The problem is relatively simple. Given the equation: (I+2A)^{-1}= \begin{bmatrix} -1 & 2 \\ 4 & 5 \end{bmatrix} Find A. My problem seems to be that I'm distributing the inverse on the LHS incorrectly. My real question then is, is the...

Cool. Thanks for the info.

It isn't real. This was one part of a practice problem I was solving as to lead me up to DEs, which I need to reacquaint myself with.

Thanks to all who responded. I believe I was doing my computations incorrectly. I am given voltage: 100e^{-500t}V And current: 20-20e^{-500t}mA Power being the product: 2e^{-1000t}(e^{500t}-1)W And derivative: -1000e^{-1000t}(e^{500t}-2) To find out when power is...

I'm trying to calculate maximum power. Once I computed power as a function of time I took its derivative and set it equal to zero. Now it need to solve for time and I can't seem to get it done. What I have is... -1000e^{-500t}-2000e^{-1000t}=0 My first inclination is to use ln, but ln(0)...

I need to solve for t and it's slipped my mind how to manipulate this. \frac{1}{4}=te^{-8t} to ln(1/4)=lnt-8t I understand the laws of logs (I think), but I still can't seem to isolate the t.

From what I understand you are correct, an equation with n derivatives requires n initial conditions. Not too sure if the same applies for coupled DEs though.

In order to determine what the integrating factor is you must first normalize the equation. \frac{dy}{dx}+P(x)y=f(x) is an example of a normalized differential equation, the IF being e^{\int P(x)dx}. Does this help?

You are both absolutely correct. I was the end of my study session, there should definitely be an equation there. I'll be posting it as soon as I can get to the book. @arildno - I know this doesn't help right now, but the problem is to be solved using Laplace transforms, not undetermined...