Solve the following initial value problems?

Question

help me to do the working yea? hahha thanks alot =)

pi = 22/7 btw is it spell as pi?

Solve the following initial value problems.

d^2y/dx^2 = -4sin(2x-pi/2) , y'(0) = 100 , y(0) = 0

and the answer is : y = - cos(2x) + 100x + 1

Answer 1

d^2y/dx^2  = -4sin(2x-pi/2)

sin(-x)  = - sinx

-sin(2x-pi/2)  = sin(pi/2-2x) =  cos2x

d^2y/dx^2  = cos2x

integrating  we get

dy/dx  =  2sin2x+C1

integrating   again we get

y = -1cos2x +C1x+C2

y'(0) = 100 = 2 sin 0 +C1

  C1  = 100

y(0)  = 0 =  - 1+100 * 0 +C2

C2-1  = 0   ==> C2 = 1

y = -cos2x+100x+1

Answer 2

d²y/dx² = -4sin(2x-π/2)

Integrate this once to get:

dy/dx = 2cos(2x-π/2) + c1

(Don't forget the constant of integration)

When x=0, dy/dx = 100

2cos(-π/2) + c1 = 100

0 + c1 = 100

c1 = 100

dy/dx = 2cos(2x-π/2) + 100

Integrate this to get

y = sin(2x-π/2) + 100x + c2

(Don't forget the constant of integration - again)

sin(2x-π/2)

= -sin(π/2-2x)

= -cos(2x)

y = -cos(2x) + 100x + c2

y(0) = 0

0 = -1 + 100×0 + c1

0 = -1 + c1

c1 = 1

y = -cos(2x) + 100x + 1