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    Initial value - find the solution to initial-value problem

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    Find the solution to initial-value problem:
    (x^2+1)y'= xe^y with initial condition y(0)= -1.

    © BrainMass Inc. brainmass.com December 24, 2021, 7:28 pm ad1c9bdddf
    https://brainmass.com/math/basic-calculus/initial-value-solution-initial-value-problem-192221

    SOLUTION This solution is FREE courtesy of BrainMass!

    (x^2 + 1) y' = x e^y

    (x^2 + 1) dy = x e^y dx

    dy/ e^y = x dx /(x^2 + 1)

    e^-y dy = x dx /(x^2 + 1)

    Upon integration, we get

    - e^-y = (1/2) ln (x^2 + 1) + C

    y(0) = -1

    Therefore, - e^1 = (1/2) ln 1 + C

    Thus, C = -e

    Therefore, - e^-y = (1/2) ln (x^2 + 1) - e

    e^-y = e - ln [sqrt(x^2 + 1)]

    -y = ln [e - ln{sqrt(x^2 + 1)}]

    y = - ln [e - ln{sqrt(x^2 + 1)}] is the solution.

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

    © BrainMass Inc. brainmass.com December 24, 2021, 7:28 pm ad1c9bdddf>
    https://brainmass.com/math/basic-calculus/initial-value-solution-initial-value-problem-192221

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