Integration by substitution?

Question

Solve the initial value problem.

(d^2y)/(dx^2) = 12(sec(6x))^2 tan(6x), y'(0) = 3, y(0) = -7

I am more interested in finding out the method than the answer. Thanks!

Answer 1

d²y/dx² = 12 sec²(6x) tan(6x)

y' = ∫12 sec(6x) sec(6x) tan(6x)  dx

let u = sec(6x)

du = 6 sec(6x) tan(6x) dx

sec(6x) tan(6x) = (1/6) du

sec(6x) = (1/6) du

y' = ∫12(1/6) u du

y' = ∫2 u du

y' = u² + C

y' = sec²(6x) + C

given y'(0) = 3

3 = sec(6*0)² + C

3 = 1 + C

C = 2

y' = sec²(6x) + 2

y = ∫sec²(6x) + 2

y = (1/6) tan(6x) + 2x + C

given y(0) = -7

-7 = (1/6) tan(6*0) + 2(0) + C

C = -7

y = (1/6)tan(6x) + 2x - 7