Sunday
March 26, 2017

Post a New Question

Posted by on .

Looking for some guidance please, I need to find the initial-value problem for dy/dx = 1+2cos^2x / y. y = 1 when x = 0.
any help is much appricated

  • maths - ,

    the hard part is to integrate the cos^2x

    we know that cos 2x = 2cos^2x - 1
    so replace the 2cos^2x with cos2x + 1

    then
    dy/dx = 1+2cos^2x
    = 1 + cos2x + 1 = 2 + cos2x

    then y = 2x + (1/2)sin2x + c
    given: when x=0, y = 1
    1 = 0 + sin0 + c
    c = 1

    y = 2x + (1/2)sin2x + 1

Answer This Question

First Name:
School Subject:
Answer:

Related Questions

More Related Questions

Post a New Question