Question

# A dynamic system is represented by a second order linear differential equation. 2frac{d^2x}{dt^2}+5frac{dx}{dt}-3x=0 The initial conditions are given as: when t=0, x=4 and frac{dx}{dt}=9 Solve the differential equation and obtain the output of the system x(t) as afunction of t.

Second order linear equations
A dynamic system is represented by a second order linear differential equation.
$$2\frac{d^2x}{dt^2}+5\frac{dx}{dt}-3x=0$$
The initial conditions are given as:
when $$t=0,\ x=4$$ and $$\frac{dx}{dt}=9$$
Solve the differential equation and obtain the output of the system x(t) as afunction of t.

2021-02-17
Solution. Given $$2\frac{d^2x}{dt^2}+5\frac{dx}{dt}-3x=0$$ (1)
when $$t=0$$ then $$x=4$$ and $$\frac{dx}{dt}=9$$
Auxiliary equation
$$2m^2+5m-3=0$$
$$2m^2+6m-m-3=0$$
$$(2m-1)(m+3)=0$$
$$m_1=-3$$ and $$m_2=\frac{1}{2}$$
Auxiliary equation
Then solution $$x(t)=c_1e^{3t}+c_2e^{\frac{1}{2t}}$$ (2)
$$\frac{dx}{dt}=3c_1e^{3t}+\frac{1}{2c}_2e^{\frac{t}{2}}$$
$$x(0)=4$$
$$c_1+c_2=4$$ (3)
Also $$\frac{dx}{dt}=9$$ when $$t=0$$
$$3c_1+\frac{c_2}{2}=9$$
$$6c_1+c_2=18$$ (4)
Solving (3) and (4) we get
$$5c_1=14\Rightarrow c_1=\frac{14}{5}$$
And $$c_2=\frac{6}{5}$$
Then solution (2) will becomes.
$$x(t)=\frac{14}{5}e^{3t}+\frac{6}{5}e^{\frac{t}{2}}$$