# Math Help - General solution...

1. ## 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..

2. 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?

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

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

$9\frac{d^2x}{dt^2}+6\frac{dx}{dt}+x=0$ where $\frac
{d^2x}{dt^2}$
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

4. oo and with the other intial condition x''(-3)=1/2

5. 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

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

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

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

Now we can use IC to find the constants

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

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

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

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