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" -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
$\displaystyle y\Rightarrow f(x)=c_1e^x + c_2e^{3x}$
$\displaystyle -1 \Rightarrow f(0) = c_1 + c_2$
$\displaystyle f'(x)=c_1e^x + 3c_2e^{3x}$
$\displaystyle 3 \Rightarrow f'(0) = c_1 + 3c_2$
You now have a system of equations like so:
$\displaystyle c_1 + 3c_2 = 3$
$\displaystyle c_1 + c_2 = -1$
I'm sure you can finish this!