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17 January, 13:29

Diff Eq dy/dx = (ycos (x)) / (1+y^2) initial condition is y (0) = 1 work below ... ?

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  1. 17 January, 13:47
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    Dy/dx = (ycos (x)) / (1 + y²)

    (1 + y²) / y dy = cos (x) dx

    (1/y + y) dy = cos (x) dx

    Integrating:

    ln (y) + y²/2 = sin (x) + c

    ln (1) + 1/2 = sin (0) + c

    c = 1/2

    Thus,

    ln (y) + y²/2 = sin (x) + 1/2
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