Stuck on a 2nd order linear differential equation

[jumbogala]

Homework Statement

y'' - 2y'+ 6y = 0

y(0) = 3
y(5) = 7

Find a solution y(t).

Homework Equations

The Attempt at a Solution

I found the characteristic equation: x² - 9x = 0, which has roots at 0 and 9.

Therefore y(t) = C1e^0x + C2e^9x

Using the initial conditions to solve this:
3 = C1 + C2

7 = C1 + C2(e^9)

And solving the system of equations gives
C1 = 3
C2 = 4.937

Therefore y(t) = 3 + 4.937e^9.

But this isn't the correct answer... where did I go wrong? It seems like it should work =\

 

[Dick]

Why is it that if you take the C1 and C2 you found, that C1+C2 isn't equal to 3?

 

[Office_Shredder]

You solved your system of equations incorrectly

3 = C1 + C2

7 = C1 + C2(e^9)

If C₁=3, then the first equation says C₂=0

 

[Mark44]

Therefore y(t) = 3 + 4.937e^9.

Some questions have already been raised about the constants, but also, your solution should be a function of t.

 

[jumbogala]

Whoops, somehow my C2 is off by a factor of 10000. Silly mistake... thanks! And yeah, I changed the x's to t's.