Answer to Question #122441 in Differential Equations for Joseph Se

Question

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 = ____

Answer 1

 x'' + x = 0

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

Let x = e^mt be a solution

So m²e^mt+ e^mt = 0

=> e^mt(m²+1) = 0

Auxiliary equation is m²+1=0 as e^mt ≠0

=> m = ±i

So the general solution of the given differential equation is

x = (C₁ cost + C₂ sin t)e^0*t

=> x = C₁ cost + C₂ sin t

Differentiating with respect to t

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

Using the initial conditions

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

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

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

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

=> C₁+ √3C₂ = √3 -----------eq(1)

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

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

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

So the actual solution is

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