Brushing up on the basics of diff eq

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  • #1
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Homework Statement


y(x) = C1sin(2x) + C2cos(2x)

Homework Equations


y(∏/8) = 0


The Attempt at a Solution


C1(1/2)(√2) + C2(1/2)(√2) + 1 = 0


The book jumps to the equation below and I'm having trouble figuring out how they got there.
C1 + C2 = -√2
 

Answers and Replies

  • #2
SammyS
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Homework Statement


y(x) = C1sin(2x) + C2cos(2x)

Homework Equations


y(∏/8) = 0


The Attempt at a Solution


C1(1/2)(√2) + C2(1/2)(√2) + 1 = 0


The book jumps to the equation below and I'm having trouble figuring out how they got there.
C1 + C2 = -√2
What is the dif. eq. you're trying to solve?

Multiply your "attempt at a solution" by √2 .
 
  • #3
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Determine C1 and C2 so that y(x) = C1sin(2x) + C2cos(2x) will satisfy the initial conditions y(∏/8) = 0 and y'(∏/8) = √2
 
  • #4
SammyS
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Multiply your "attempt at a solution" in your original post by √2 .
 
  • #5
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I guess it's safe to assume that C1/2 = C1 and C2/2 = C2 considering they're both unknowns at this point?
 
  • #6
SammyS
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I guess it's safe to assume that C1/2 = C1 and C2/2 = C2 considering they're both unknowns at this point?
That doesn't make sense, unless you are asking if you can replace C1/2 by C1 and C2/2 by C2 .

If [itex]C_1+C_2=-\sqrt{2}\,,[/itex] then you can replace [itex]C_2[/itex] with [itex]-C_1-\sqrt{2}[/itex]
 

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