Then suppose for y''-9y'+14y my yp=Ax2e4x
Then
y'=2Axe4x+4Ax2e4x
y''=2Ae4x+8Axe4x+8Axe4x+16Ax2e4x
When you plug this all back into the original equation, equate the coefficients to solve for A, it doesn't work. Need more than just one constant.
I don't quite understand why your guess for the particular solution has the form Ax2e4x as opposed to (Ax2+Bx+C)e4x. Shouldn't you use the general form of the second degree polynomial?
There are two questions that I am trying to solve on web assignment. The goal is to find a general form of a particular solution to each ODE. The question asks me to represent all constants in the solution using "P,Q,R,S,T..etc.", in that order.
1. y'''-9y''+14y'=x2
2. y''-9y'+14y=x2e4x...
When I solve for Isc through a and b using Norton and Thevenin equivalents between c and d I get different results. Why is that?
Norton =0.5mA
Thevenin =2mA
Hang on. I see that R4 directly connects E and F (due to the short circuit), but how can any current going from E to F passing through R3 NOT also pass through some other resistor?
Because there are nodes in between all resistors, it is almost impossible simplify the circuit using parallel/series connection. Should I redraw the circuit in some way?
Homework Statement
I got this on a quiz today - Find the equivalent resistance between E and F.
Now, the problem with this question is that I get TWO different answers when I trace from:
1. e to f
2. f to e
I know one of them is definitely wrong (or maybe both are?), but I don't...
Sorry, I haven't quite learned that yet. Do you mind elaborating a little? Is that another technique for solving differential equations? Something called substitution, maybe?
Thank you