General solution...

• May 28th 2009, 06:48 PM
Raidan
General solution...
can someone help me get the general solution for this..I get confused about the derivative of x

9d^2x/dt^2+6dx/dt+x=0
,X(-3)=2 ,dx/dt(-3)=1/2

I get x(t)=(C1+C2t)e^-1/3t

I cannot seem to get the answers for the initial conditions above..
• May 28th 2009, 08:38 PM
VonNemo19
Quote:

Originally Posted by Raidan
can someone help me get the general solution for this..I get confused about the derivative of x

9d^2x/dt^2+6dx/dt+x=0 ,X(-3)=2 ,dx/dt(-3)=1/2

I get x(t)=(C1+C2t)e^-1/3t

I cannot seem to get the answers for the initial conditions above..

I'm not sure that I understand. Is this what you have?

$\displaystyle 9\frac{d^2x}{dt^2}+6\frac{dx}{dt}+x=0$ where $\displaystyle \frac {d^2x}{dt^2}$ is the second derivative of x with respect to t; and x is a function of t and x(-3)=2?
• May 28th 2009, 09:02 PM
Raidan
Quote:

Originally Posted by VonNemo19
I'm not sure that I understand. Is this what you have?

$\displaystyle 9\frac{d^2x}{dt^2}+6\frac{dx}{dt}+x=0$ where $\displaystyle \frac {d^2x}{dt^2}$ is the second derivative of x with respect to t; and x is a function of t and x(-3)=2?

yes thats what I mean..hw would i do this
• May 28th 2009, 09:03 PM
Raidan
oo and with the other intial condition x''(-3)=1/2
• May 28th 2009, 10:14 PM
TheEmptySet
Quote:

Originally Posted by Raidan
can someone help me get the general solution for this..I get confused about the derivative of x

9d^2x/dt^2+6dx/dt+x=0 ,X(-3)=2 ,dx/dt(-3)=1/2

I get x(t)=(C1+C2t)e^-1/3t

I cannot seem to get the answers for the initial conditions above..

So this is a 2nd order linear ODE with constant coeffeints so the auxillary equaiton is

$\displaystyle 9m^2-6m+1=0 \iff (3m-1)(3m-1)=0 \iff (3m-1)^2=0$

So we have one repeated root. So we multiply by t to get the 2nd linearly independant solution.

$\displaystyle x(t)=c_1e^{\frac{t}{3}}+c_2te^{\frac{t}{3}}$

Now we can use IC to find the constants

$\displaystyle x'(t)=\frac{c_1}{3}e^{\frac{t}{3}}+\frac{c_2}{3}te ^{\frac{t}{3}}+c_2e^{\frac{t}{3}}$

$\displaystyle x'(-3)=\frac{1}{2}=\frac{c_1}{3}e^{-1}+\frac{c_2}{3}(-3)e^{-1}+c_2e^{-1} \iff 3e=2c_1-6c_2+6c_2 \iff c_1=\frac{3e}{2}$

$\displaystyle x(-3)=2=\frac{3e}{2}e^{-1}+c_2(-3)e^{-1} \iff 4e=3e-6c_2\iff c_2=-\frac{e}{6}$

$\displaystyle x(t)=\left( \frac{3e}{2}\right)e^{\frac{t}{3}}-\frac{e}{6}te^{\frac{t}{3}}$