Ask Question
14 May, 06:27

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

+5
Answers (1)
  1. 14 May, 06:56
    0
    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)
Know the Answer?
Not Sure About the Answer?
Get an answer to your question ✅ “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 ...” in 📙 Mathematics if there is no answer or all answers are wrong, use a search bar and try to find the answer among similar questions.
Search for Other Answers