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

    © BrainMass Inc. brainmass.com December 24, 2021, 4:43 pm ad1c9bdddf
    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.

    This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

    © BrainMass Inc. brainmass.com December 24, 2021, 4:43 pm ad1c9bdddf>
    https://brainmass.com/math/integrals/integral-initial-value-problems-4221

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