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/2) = 0, x' (π/2

Incomplete Question

The value of x' (π/2) is not given.

I'll assume x' (π/2) to be 1

Answer:

x (t) = cos (t)

Step-by-step explanation:

Given

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

x (π/2) = 0

x' (π/2) = 1

First, we'll differentiate x (t) to give x' (t)

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

Given that x (π/2) = 0

We substitute π/2 for t in x (t)

So,

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

c1 cos (π/2) + c2 sin (π/2) = 0

cos (π/2) = 0 and sin (π/2) = 1;

So, we have

c1 * 0 + c2 * 1 = 0

0 + c2 = 0

c2 = 0.

Given that x' (π/2) = 1

We substitute π/2 for t in x' (t).

So,

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

-c1 * sin (π/2) + c2 * cos (π/2) = 1

-c1 * 1 + c2 * 0 = 1

-c1 = + 0 = 1

-c1 = 1

c1 = 1.

So,

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

=>

x (t) = 1 * cos (t) + 0 * sin (t)

x (t) = cos (t)