O.D.E. Complementary and Particular Solution

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The discussion centers on solving the ordinary differential equation (ODE) given by y'' - 2y' - 3y = 6. The complementary solution is identified as y_c = C_1 * e^(-x) + C_2 * e^(3x), while the particular solution is y_p = -2. The user initially calculates C_1 and C_2 as 2 and 3, respectively, but the correct values are C_1 = 1 and C_2 = 4, as stated in the reference book. The discrepancy was attributed to a simple arithmetic error in the user's calculations.

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kahless2005
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Not exactly homework, but it is a problem I'm having...

Im given an ode that reads:
y"-2y'-3y = 6;
[itex]y_c = C_1 * /exp^-x + C_2 * /exp^3x[/itex]
[itex]y_p[/itex] is -2

y(0) = 3
y'(0) = 11

Now I am tasked to find what [itex]C_1[/itex] and [itex]C_2[/itex] are.

I know that y(x) = [itex]y_c + y_p[/itex]
so:
y = [itex]C_1 * /exp^-x + C_2 * /exp^3x[/itex] - 2
and
y' = [itex]-C_1 * /exp^-x + 3 * C_2 * /exp^3x[/itex]

The book defines the answers as:
[itex]C_1[/itex] = 1 and [itex]C_2[/itex] = 4

Yet when I work it out, I've gotten [itex]C_1[/itex] = 2 and [itex]C_2[/itex] = 3.

What am I doing wrong?

NOTE: I hope I did the itex right... my computer isn't showing them... :biggrin:
 
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Nevermind... I got it to work. I had a simple Aritmatic Error
 

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