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- AmTran_Bus

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- AmTran_Bus

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- AmTran_Bus

|dw:1440809128053:dw|

- AmTran_Bus

So I do not have a clue what to do about the boundary equation.

- AmTran_Bus

But I can show you how I solved the original part.

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- tkhunny

I'd try something like \(y(x) = Ae^{2x} + Be^{-2x} + Cx + D\)
Or, I might notice that \(r^{2} - 4 = 0\) has solutions r = 2 and r = -2 and then forget about the C and D terms.

- AmTran_Bus

|dw:1440809556496:dw|
is what I got. But I have no clue what a boundary is nor how to apply it.

- AmTran_Bus

and the book says that is right.

- tkhunny

?? Substitute x = 0
\(c_{1}e^{2(0)} + c_{2}e^{-2(0)} = 2\)
Now, find the first derivative and do it again. This gives two equations in two unknowns and you are nearly done.

- AmTran_Bus

So you gotta take deriv. and then substitute.

- tkhunny

If you want to use the second boundary condition, yes.

- AmTran_Bus

Ohhhhhh. That is where I am messing up.

- AmTran_Bus

Duh duh me. Thanks so much.

- AmTran_Bus

So 2=c1+c2 and 2= c1-c2

- tkhunny

Shedding a little light is very rewarding.

- AmTran_Bus

Sure nuff. Especially since this is pchem.

- AmTran_Bus

Hum. So how where does the 2e^(2x) come from at the end @tkhunny ? I know 4=2c1 and so c1=2 and c2=0 so does that 2 just apply to both the coeff. and the e^whatever?

- tkhunny

"pchem"?! You have my sympathies. You CAN get through it.

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