Answer to Question #136933 in Differential Equations for shucayb ali jama

Question

d2^x/dt^2 - 6 dx/dt +9x =0; when t =0, x =2 and dx/dt = 0

Answer 1

$\frac{d^2x}{dt^2}-6\frac{dx}{dt}+9x=0$, $x(0)=2$ and $\frac{dx(0)}{dt}=0$.

Let us solve the characteristic equation:

$k^2-6k+9=0$

$(k-3)^2=0$

$k_1=k_2=3$

Therefore, the general solution is of the following form:

$x(t)=e^{3t}(C_1+C_2t)$

Then $\frac{dx}{dt}=3e^{3t}(C_1+C_2t)+C_2e^{3t}=e^{3t}(3C_1+3C_2t+C_2)$

Taking into account that $x(0)=2$ and $\frac{dx(0)}{dt}=0$ we conclude that $C_1=2$ and $3C_1+C_2=0$. Thus $C_2=-3C_1=-6$.

Consequently, the particular solution is the following:

$x(t)=e^{3t}(2-6t)$