AmTran_Bus
  • AmTran_Bus
Need DE help!
Mathematics
jamiebookeater
  • jamiebookeater
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AmTran_Bus
  • AmTran_Bus
|dw:1440809128053:dw|
AmTran_Bus
  • AmTran_Bus
So I do not have a clue what to do about the boundary equation.
AmTran_Bus
  • AmTran_Bus
But I can show you how I solved the original part.

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tkhunny
  • 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
  • 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
  • AmTran_Bus
and the book says that is right.
tkhunny
  • 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
  • AmTran_Bus
So you gotta take deriv. and then substitute.
tkhunny
  • tkhunny
If you want to use the second boundary condition, yes.
AmTran_Bus
  • AmTran_Bus
Ohhhhhh. That is where I am messing up.
AmTran_Bus
  • AmTran_Bus
Duh duh me. Thanks so much.
AmTran_Bus
  • AmTran_Bus
So 2=c1+c2 and 2= c1-c2
tkhunny
  • tkhunny
Shedding a little light is very rewarding.
AmTran_Bus
  • AmTran_Bus
Sure nuff. Especially since this is pchem.
AmTran_Bus
  • 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
  • tkhunny
"pchem"?! You have my sympathies. You CAN get through it.

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