# 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.

A dynamic system is represented by a second order linear differential equation.
$2\frac{{d}^{2}x}{d{t}^{2}}+5\frac{dx}{dt}-3x=0$
The initial conditions are given as:
when and $\frac{dx}{dt}=9$
Solve the differential equation and obtain the output of the system x(t) as afunction of t.
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Solution. Given $2\frac{{d}^{2}x}{d{t}^{2}}+5\frac{dx}{dt}-3x=0$ (1)
when $t=0$ then $x=4$ and $\frac{dx}{dt}=9$
Auxiliary equation
$2{m}^{2}+5m-3=0$
$2{m}^{2}+6m-m-3=0$
$\left(2m-1\right)\left(m+3\right)=0$
${m}_{1}=-3$ and ${m}_{2}=\frac{1}{2}$
Auxiliary equation
Then solution $x\left(t\right)={c}_{1}{e}^{3t}+{c}_{2}{e}^{\frac{1}{2t}}$ (2)
$\frac{dx}{dt}=3{c}_{1}{e}^{3t}+{\frac{1}{2c}}_{2}{e}^{\frac{t}{2}}$
$x\left(0\right)=4$
${c}_{1}+{c}_{2}=4$ (3)
Also $\frac{dx}{dt}=9$ when $t=0$
$3{c}_{1}+\frac{{c}_{2}}{2}=9$
$6{c}_{1}+{c}_{2}=18$ (4)
Solving (3) and (4) we get
$5{c}_{1}=14⇒{c}_{1}=\frac{14}{5}$
And ${c}_{2}=\frac{6}{5}$
Then solution (2) will becomes.
$x\left(t\right)=\frac{14}{5}{e}^{3t}+\frac{6}{5}{e}^{\frac{t}{2}}$