x'' + x = 0
Let x = emt be a solution
So m²emt+ emt = 0
=> emt(m²+1) = 0
Auxiliary equation is m²+1=0 as emt ≠0
=> m = ±i
So the general solution of the given differential equation is
x = (C1 cost + C2 sin t)e0*t
=> x = C1 cost + C2 sin t
Differentiating with respect to t
= - C1 sin t + C2 cos t
Using the initial conditions
x = when t =
= 0 when t =
So C1 cos( ) + C2 sin( ) = √3/2
=> C1 + C2 =
=> C1+ √3C2 = √3 -----------eq(1)
Again = 0 when t =
So,. - C1 sin( ) + C2 cos( ) = 0
=> - C1 + C2 = 0
=> -√3C1 + C2 = 0
Multiplying by √3
-3C1 + √3C2 = 0 --------------eq(2)
Subtracting eq(2) from eq(1) we get
4C1 = √3
So C1 = and hence C2 = √3C1=
So the actual solution is
x = cos t + sin t
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