Verify that the basis to the general solution y1=e^(-5x) and y2=e^(6x) , y''-y'-30y=0
To verify basis, the first step is to take the first and second derivative of y1, which will give you 25e^(-5x), and then plug it back into the differential equation, thus 25e^(-5x)-(-5e^(-5x))-30(e^(-5x)) = 0
What is the limit of the function f(x)= sin(2x)/x ?
Using the identity lim --> 0 sin(x)/x is equal 1, multiply the function f(x) by 2/2, this in turn becomes 2sin(2x)/(2x), using the identity, sin(2x)/(2x) is equal to 1, thus the only thing left is 2x1 which is 2.
What is a capacitors' impedance and behavior with a high frequency input signal?
With a high frequency AC input signal, the carriers produce an electromagnetic force that allows the signal to pass through, the capacitor acts like a short, thus allowing the signal to pass; which unlike a DC signal, it acts as an open, which blocks the DC signal (zero frequency).