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Solving equation simultaneously for constant

  • #1
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This is for 2nd order homogenous differential equation but that isn't what needs to be solved. What I need help with is solving simultaneously-- something that I am terrible with.

[itex]y=c_1e^t+c_2e^{-t}[/itex]
With t = 0, y = 2 we have [itex]c_1 + c_2 = 2[/itex]

Upon differentiation of y, we obtain: [itex]y'=c_1e^t-c_2e^{-t}[/itex]

With t = 0, y' = -1 [itex]c_1 - c_2 = -1[/itex]

Now the book says that upon solving these two equations simultaneously, [itex] c_1 = 1/2 & c_2 = 3/2[/itex]

So I try to verify the result, I set up the two equations to solve by elimination

[itex]c_1 + c_2 = 2[/itex]
[itex]c_1 - c_2 = -1[/itex]
c_2 cancels out & we obtain c_1 = 1 which is the wrong conclusion. I can reason out that c_1 should be 1/2 & c_2 should be 3/2, but first and foremost I want to know why solving by elimination gives the wrong answer and what method would be the correct one.
 

Answers and Replies

  • #2
SammyS
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This is for 2nd order homogenous differential equation but that isn't what needs to be solved. What I need help with is solving simultaneously-- something that I am terrible with.

[itex]y=c_1e^t+c_2e^{-t}[/itex]
With t = 0, y = 2 we have [itex]c_1 + c_2 = 2[/itex]

Upon differentiation of y, we obtain: [itex]y'=c_1e^t-c_2e^{-t}[/itex]

With t = 0, y' = -1 [itex]c_1 - c_2 = -1[/itex]

Now the book says that upon solving these two equations simultaneously, [itex] c_1 = 1/2 \text{ & } c_2 = 3/2[/itex]

So I try to verify the result, I set up the two equations to solve by elimination

[itex]c_1 + c_2 = 2[/itex]
[itex]c_1 - c_2 = -1[/itex]
c_2 cancels out & we obtain c_1 = 1 which is the wrong conclusion. I can reason out that c_1 should be 1/2 & c_2 should be 3/2, but first and foremost I want to know why solving by elimination gives the wrong answer and what method would be the correct one.
If
[itex]c_1 + c_2 = 2[/itex]
[itex]c_1 - c_2 = -1[/itex]​
Then adding the two equations gives
[itex]2c_1=1\ [/itex]​
 
  • #3
1,291
0
If
[itex]c_1 + c_2 = 2[/itex]
[itex]c_1 - c_2 = -1[/itex]​
Then adding the two equations gives
[itex]2c_1=1\ [/itex]​
Oh--duh. That was probably the easiest problem you've had to solve to date :rofl: .

1 + 1 ≠ 1 .. maybe it is time I get some sleep. :grumpy:
 

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