Question

Equation is x"-4x=0, general solution is x(t) = C1e^-2t+C2e^2t, initial condition is x(0) = 1, x'(0)=0

Answer 1

Answer:

$x(t)=\frac{1}{2}e^{-2t}+\frac{1}{2}e^{2t}$

Step-by-step explanation:

The given differential equation is $x''-4x=0$

The characteristics equation is given by

$r^2-4=0\\\\r^2=4\\\\r=\pm2$

Therefore, the solution of the DE is given by

$x(t)=c_1e^{-2t}+c_2e^{2t}$

On differentiating, we get

$x'(t)=-2c_1e^{-2t}+2c_2e^{2t}$

Apply the initial conditions

x(0)=1

$1=c_1e^{0}+c_2e^{0}\\\\c_1+c_2=1...(i)$

Second condition is x'(0)=0

$0=-2c_1e^{0}+2c_2e^{0}\\\\-2c_1+2c_2=0\\\\c_1-c_2=0...(ii)$

Add (i) and (ii)

$2c_1=1\\\\c_1=\frac{1}{2}$

Substituting this value in (ii)

$\frac{1}{2}-c_2=0\\\\c_2=\frac{1}{2}$

Hence, the solution of the DE is

$x(t)=\frac{1}{2}e^{-2t}+\frac{1}{2}e^{2t}$