Question #122441
1. In this problem, x = c1 cos t + c2 sin t is a two-parameter family of solutions of the second-order DE x'' + x = 0. Find a solution of the second-order IVP consisting of this differential equation and the given initial conditions.

x(pi/3) = square root of 3 / 2, x prime (pi/3) = 0

x = ____
1
Expert's answer
2020-06-16T15:54:40-0400

 x'' + x = 0

d2xdt2+x=0\frac {d^2x}{dt^2} + 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

dxdt\frac {dx}{dt} = - C1 sin t + C2 cos t

Using the initial conditions

x = 32\frac {√3}{2} when t = π3\frac {π}{3}

dxdt\frac {dx}{dt} = 0 when t = π3\frac {π}{3}

So C1 cos(π3\frac {π}{3} ) + C2 sin(π3\frac {π}{3} ) = √3/2

=> 12\frac {1}{2} C1 + 32\frac {√3}{2} C2 = 32\frac {√3}{2}

=> C1+ √3C2 = √3 -----------eq(1)

Again dxdt\frac {dx}{dt} = 0 when t = π3\frac {π}{3}

So,. - C1 sin(π3\frac {π}{3} ) + C2 cos(π3\frac {π}{3} ) = 0

=> - 32\frac {√3}{2} C1 + 12\frac {1}{2} 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 = 34\frac {√3}{4} and hence C2 = √3C1= 34\frac {3}{4}

So the actual solution is

x = 34\frac {√3}{4} cos t + 34\frac {3}{4} sin t




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