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=0X(-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..

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- May 28th 2009, 06:48 PMRaidanGeneral 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 PMVonNemo19
- May 28th 2009, 09:02 PMRaidan
- May 28th 2009, 09:03 PMRaidan
**oo and with the other intial condition x''(-3)=1/2** - May 28th 2009, 10:14 PMTheEmptySet
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

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

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

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

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