Solving equation simultaneously for constant

[Nano-Passion]

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.

$y=c_1e^t+c_2e^{-t}$
With t = 0, y = 2 we have $c_1 + c_2 = 2$

Upon differentiation of y, we obtain: $y'=c_1e^t-c_2e^{-t}$

With t = 0, y' = -1 $c_1 - c_2 = -1$

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

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

$c_1 + c_2 = 2$
$c_1 - c_2 = -1$
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.

 

[SammyS]

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.

$y=c_1e^t+c_2e^{-t}$
With t = 0, y = 2 we have $c_1 + c_2 = 2$

Upon differentiation of y, we obtain: $y'=c_1e^t-c_2e^{-t}$

With t = 0, y' = -1 $c_1 - c_2 = -1$

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

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

$c_1 + c_2 = 2$
$c_1 - c_2 = -1$
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

$c_1 + c_2 = 2$
$c_1 - c_2 = -1$​

Then adding the two equations gives

$2c_1=1\$​

 

[Nano-Passion]

If

$c_1 + c_2 = 2$
$c_1 - c_2 = -1$​

Then adding the two equations gives

$2c_1=1\$​

Oh--duh. That was probably the easiest problem you've had to solve to date .

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