# Thread: Constant coefficients initial value problem

1. ## Constant coefficients initial value problem

y" -4y' + 3y = 0 y(0)= -1 ; y'(0)= 3

Here's how I tried to solve it:
1. r^2 - 4r + 3 = 0
2. (r-1)(r-3) = 0
3. r= 1, 3
4. Thus y=c1(e^x) + c2(e^3x)

But now I'm stuck. How do I use the initial values to solve for c1, c2?

Thanks,

Jim

2. Originally Posted by Jim Newt
y" -4y' + 3y = 0 y(0)= -1 ; y'(0)= 3

Here's how I tried to solve it:
1. r^2 - 4r + 3 = 0
2. (r-1)(r-3) = 0
3. r= 1, 3
4. Thus y=c1(e^x) + c2(e^3x)

But now I'm stuck. How do I use the initial values to solve for c1, c2?

Thanks,

Jim
$y\Rightarrow f(x)=c_1e^x + c_2e^{3x}$

$-1 \Rightarrow f(0) = c_1 + c_2$

$f'(x)=c_1e^x + 3c_2e^{3x}$

$3 \Rightarrow f'(0) = c_1 + 3c_2$

You now have a system of equations like so:

$c_1 + 3c_2 = 3$
$c_1 + c_2 = -1$

I'm sure you can finish this!