Question

The family functions y=c1*e^x*cos(x)+c2*e^x*sin(x)
is a general solution of y''-2y'+2y=0

Decide whether there is a member of the family that satisfies the boundary condition y(-pi/2)=0 and y(pi/2)=0

Answer 1

we want to look at the possible values of the constants c1 and c2

plug in the values into the equation and you will get a system
if the system is solvable for a unique solution of c1 and c2, then that is your answer

Answer 2

I"ve worked it out, and obviously the cos(x) side of the equation is irrelevant. You end up with two equations 0=c2*e^x*sin(-pi/2) and 0=c2*e^x*cos(-pi/2). Works out to 0= -xe^(-pi/2) and 0=xe^(pi/2)

Answer 3

I'm having problem expressing it in the form y=(some equation here). I assume there are infinite solutions based a single parameter C. I just can't find what that value is

Answer 4

\[y(-\frac{\pi}2)=-c_1e^{-\frac\pi2}\cos(-\frac{\pi}2)+c_2e^{-\frac{\pi}2}\sin(-\frac{\pi}2)=-c_2e^{-\frac\pi2}=0\]\[y(-\frac{\pi}2)=-c_1e^{-\frac\pi2}\cos(-\frac{\pi}2)+c_2e^{\frac{\pi}2}\sin(\frac{\pi}2)=c_2e^{\frac\pi2}=0\]so it looks like c2 is trying to be zero, which I'm pretty sure means there is no such equation in the family.

Answer 5

oh that's all supposed to be positive pi/2 in the second line

Answer 6

\[y(-\frac{\pi}2)=-c_1e^{-\frac\pi2}\cos(-\frac{\pi}2)+c_2e^{-\frac{\pi}2}\sin(-\frac{\pi}2)=-c_2e^{-\frac\pi2}=0\]\[y(\frac{\pi}2)=-c_1e^{\frac\pi2}\cos(\frac{\pi}2)+c_2e^{\frac{\pi}2}\sin(\frac{\pi}2)=c_2e^{\frac\pi2}=0\]

Answer 7

hm... you make a point
I'm not sure, let me think about it
or perhaps @JamesJ can help

Answer 8

I thought it was 0 at first two, but my online homework disagrees :) thanks for the insight though

Answer 9

Right. So c1 can be anything and c2 = 0

Answer 10

Cool, nice and simple
thanks James, and that feature is pretty cool :D

Answer 11

Yes, it is. I just hope no every man and his dog can mention me!

Answer 12

Also good, my question is how would I write that in terms of y=______ , though I agree your answer is much more intuitive

Answer 13

*not

Answer 14

I hear you
brycoop: I believe you can write
if you want to express that you can write\[y=te^x\cos x;t\in \mathbb R\]

Answer 15

y=c1*e^x*cos(x)

Answer 16

or yeah, same thing...

Answer 17

ehh webworks won't take it, which doesn't mean its wrong, I'll email my prof and see what she says, thanks a bunch guys

Answer 18

perhaps your webwork would be happier if you chose a random number for c, like c=1
have you tried that?

Answer 19

just needed to add a * forget how picky webworks can be with multiplication. Try to avoid being implicit in the future I suppose haha