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.

Question
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.

Answers (1)

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.
\(x(t)=\frac{14}{5}e^{3t}+\frac{6}{5}e^{\frac{t}{2}}\)
0

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