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(π/3) = 3 2 , x'(π/3) = 0

solution:

we have the solution as

x(t)= c1*cos(t) + c2*sin(t)

=>x'(t) = -c1*sin(t) +c2*cos(t)

given x(π/3)=sqrt(3)/2 and x'(π/3)=0

=>sqrt(3)/2= c1*(1/2) + c2*(sqrt(3)/2) > (1)

and 0= -c1*(sqrt(3)/2) + c2*(1/2) >(2)

=>c2=c1*sqrt(3)

using this in (1) we get

sqrt(3)/2= c1*(1/2) + c1*(3/2)

=>c1= sqrt(3)/4

and c2=3/4

x(t) = (sqrt(3)/4)*cos(t) +(3/4)*sin(t)

you would need to know all the numbers that are on there

step-by-step explanation: