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Question

The solution of differential equation d2Ψdt2+6dΨdt+13Ψ=0 is, given Ψ(0)=0 is,

A
c.e3t[cos 2t]
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B
c.e3t[sin 2t]
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C
e3t[c1 cos 2t+c2 sin 2t](c1,c20)
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D
e3t[c1 cos 2t+c2 sin 2t](c1,c20)
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Solution

The correct option is B c.e3t[sin 2t]
Auxiliary equations,

D2+6D+13=0
i.e m2+6m+13=0

m=3±2i

Solution of DE is Ψ=e3t[c1 cos2t+c2sin 2t]

Now, Ψ(0)=e0[c1.1+0]=0
c1=0

So,
Solution of DE is, Ψ=e3t[c2 sin 2t]
=e3t[c sin 2t]

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