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15 August, 02:33

Solve the differential equation dy dx equals the quotient of y times the cosine of x and the quantity 1 plus y squared with the initial condition y (0) = 1.

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  1. 15 August, 06:31
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    We rearrange the given differential equation to combine all the y terms on the left side of the equation and the x terms on the right side:

    [ (1 + y^2) / y] dy = cos (x) dx

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

    We integrate both sides to get

    ln y + (y^2 / 2) = sin (x) + C

    Then, we apply the initial condition y = 1 at x = 0 to get the value of C:

    ln (1) + (1^2 / 2) = sin (0) + C

    C = ln (1) + (1^2 / 2) - sin (0)

    C = 1/2

    Therefore, our final answer is

    ln (y) + (y^2 / 2) = sin (x) + 1/2
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