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..
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
$\displaystyle x'(t)=\frac{c_1}{3}e^{\frac{t}{3}}+\frac{c_2}{3}te ^{\frac{t}{3}}+c_2e^{\frac{t}{3}}$
$\displaystyle x'(-3)=\frac{1}{2}=\frac{c_1}{3}e^{-1}+\frac{c_2}{3}(-3)e^{-1}+c_2e^{-1} \iff 3e=2c_1-6c_2+6c_2 \iff c_1=\frac{3e}{2}$
$\displaystyle x(-3)=2=\frac{3e}{2}e^{-1}+c_2(-3)e^{-1} \iff 4e=3e-6c_2\iff c_2=-\frac{e}{6}$
$\displaystyle x(t)=\left( \frac{3e}{2}\right)e^{\frac{t}{3}}-\frac{e}{6}te^{\frac{t}{3}}$