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Solve the initial value problem:
d²y/dx²= sec²x
y(0)=0 and y'(0)=1

https://brainmass.com/math/integrals/integral-initial-value-problems-4221

## SOLUTION This solution is FREE courtesy of BrainMass!

d^2y/dx^2 = sec^2x

Integrating once wrt x we have;

dy/dx = tanx + c , where c is the constant of integration

Applying boundary conditions we have ;

1 = tan(0)+ C ( given that y'(o) =1 )
So C = 1 ( Coz Tan(0) is 0).

dy/dx = tanx + 1 Integrating again wrt x we have:

y = -Ln ( Cosx) + x ( Integration of tan x is - Ln(Cosx))
Applying boundary conditions ,

0 = Ln ( Cos 0) + 0 which satisfies the equation as Cos 0 =1 and Ln(1)=0.

So Initially y = -Ln(Cos(x))+ x........Final Answer.

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