For the Ordinary Differential Equation {d^{2} x}/{d t^{2}}-5 \frac{d x}{d t}+6 x=0, with initial conditions {x}(0)=0 and { {d} {x}}/{ {dt}}(0)=10, the solution is

Correct Answer is : −10e2t+10e3t

GATE Civil Engineering (CE) 2020 Shift 1 Solved Paper

Question : 43 of 65

Marks: +1, -0

For the Ordinary Differential Equation

\;{d^{2} x}/{d t^{2}}-5 \;{d x}/{d t}+6 x=0

, with initial conditions

{x}(0)=0

and

\;{ {d} {x}}/{ {dt}}(0)=10

, the solution is

Solution:

\;{d^{2} x}/{d t^{2}}-\;{5 d x}/{d t}+6 x=0

D^{2}-5 D+6=0

D^{2}-3 D-2 D+6=0

D(D-3)-2(D-3)=0

D=2,3

Solution will be,

x=C_{1} e^{3 t}+C_{2} e^{2 t}

Applying boundary condition

x(0)=0

C_{1}+C_{2}=0 →(1)

\;{d x}/{d t}=3 c_{1} e^{3 t}+2 c_{2} e^{2 t}

\;{d x}/{d t}(0)=10

10=3 c 1+2 c 2 →(2)

Form equation (1) and (2)

{C}_{1}=10 {C}_{2}=-10

So, final solution becomes,

x=10 e^{3 t}-10 e^{2 t}

x=-10 e^{2 t}+10 e^{3 t}

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