# Thread: Derivative of General Solution

1. ## Derivative of General Solution

this should be simple but I can't get it to come out right

i have the general solution as x(t)= C1e^(-t)cos4.36t + C2e^(-t)sin4.36t

can someone get the first deriv of x(t)? C1 and C2 are what I need to solve for to find the particular solution

2. is your general solution correct??

3. Originally Posted by johan
yeah i'm pretty sure..the roots to my equation were M1=-1+4.36i and M2=-1-4.36i

4. Originally Posted by chief27
this should be simple but I can't get it to come out right

i have the general solution as x(t)= C1e^(-t)cos4.36t + C2e^(-t)sin4.36t

can someone get the first deriv of x(t)? C1 and C2 are what I need to solve for to find the particular solution
Maybe you should post the whole question ....

5. Originally Posted by mr fantastic
Maybe you should post the whole question ....
The DE is y"+2y'+20x = 0 initial conditions are x(0)=1 and x'(0)=0

my characteristic equation is m^2+2m+20=0

use initial conditions to determine the particular solution

6. ## Second order differential equation

Hello chief27
Originally Posted by chief27
The DE is y"+2y'+20x = 0 initial conditions are x(0)=1 and x'(0)=0

my characteristic equation is m^2+2m+20=0

use initial conditions to determine the particular solution
I agree with your general solution. I've left the term as $\sqrt{19}$ rather than 4.36.

$x = e^{-t}(C_1\cos\sqrt{19}t+C_2\sin\sqrt{19}t)$

$\Rightarrow x(0) = C_1 = 1$

$\Rightarrow x = e^{-t}(\cos\sqrt{19}t+C_2\sin\sqrt{19}t)$

$\Rightarrow \dot{x} = e^{-t}(-\sqrt{19}\sin\sqrt{19}t + \sqrt{19}C_2\cos\sqrt{19}t) - e^{-t}(\cos\sqrt{19}t+C_2\sin\sqrt{19}t)$

$\Rightarrow \dot{x}(0) = \sqrt{19}C_2 - 1 = 0$

$\Rightarrow C_2 = \frac{1}{\sqrt{19}}$

$\Rightarrow x = e^{-t}\left(\cos\sqrt{19}t+\frac{1}{\sqrt{19}}\sin\sqr t{19}t\right)$