Solve the differential equation: dy/dx + y*sec(x)*tan(x) = e^[-sec(x)] when y(0) = 1?

Question

anybody have any ideas?

any ideas? well after i separate variables i differentiate but how?

Answer 1

This differential equation can be solved by the method of variation of parameters.

Consider a differential equation of the form:
y' + p(x) × y = q(x)
For the right hand side set to zero you get the homogeneous solution.
h(x) = c × exp{-∫ p(x) dx}
where c is the constant of integration.
To find the solution of the inhomogeneous differential equation assume c is a function of x instead of a constant.
y(x) = c(x) × h(x)
Substituting into the differential equation leads to
c'(x) × h(x) = q(x)
with the solution
c(x) = ∫ q(x) ÷ h(x) dx + C
= ∫ q(x) × exp{+∫ p(x) dx} dx + C
Therefore the general solution for such an differential equation is:
y = exp{-∫ p(x) dx} × (∫ q(x) × exp{∫ p(x) dx } dx +C)

Here
p(x) = sec(x) × tan(x)
q(x) = exp{-sec(x)}

Because
d/dx(sec(x)) = sec(x) × tan(x)
∫ sec(x) × tan(x) dx = sec(x)

y = exp{-sec(x)} × (∫ exp{-sec(x) } × exp{sec(x)}dx + C)
= exp{-sec(x)} × (∫ 1dx + C)
= exp{-sec(x)} × (x + C)

The Constant C can be calculated by applying the the boundary condition
y(0) = 1
1 = exp(-sec(0)} × (0+C)
=>
C = e

The solution is
y(x) = exp{-sec(x)} × (x+e)