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    is it possible to solve

    \frac{dy}{dt}  = \sqrt y - \sqrt x ?

    the bear

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    i have plotted a gradient field using x and y axes...

    how should the solution curves from the field be used if the formula is dy/dt = x^0.5 - y^0.5 rather than dy/dx = x^0.5 - y^0.5 ?

    the baer

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    How about we re-arrange it to: \frac{dy}{dt} - \sqrt y = - \sqrt x

    Notice how the differential equation involves only a "dy" and "dt" NOT both "dy" and "dx" so we can simply treat \sqrt x as a constant

    Now it's in the form \frac{dy}{dt} - \sqrt y = constant so we could perhaps solve it using the Integrating Factor

    I've not actually tried it out on paper mind, so it may not work. But hope it helps
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    (Original post by McMannBean)
    How about we re-arrange it to: \frac{dy}{dt} - \sqrt y = - \sqrt x

    Notice how the differential equation involves only a "dy" and "dt" NOT both "dy" and "dx" so we can simply treat \sqrt x as a constant

    Now it's in the form \frac{dy}{dt} - \sqrt y = constant so we could perhaps solve it using the Integrating Factor

    I've not actually tried it out on paper mind, so it may not work. But hope it helps
    You can't use the integrating factor method because the ODE is non-linear.
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    let's try a substitution then:

    Let U = sqrt(y)

    so we end up with 2U*dU/dt - U = const

    => dU/dt - const*U^-1 = 1

    This should be solveable
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    (Original post by McMannBean)
    let's try a substitution then:

    Let U = sqrt(y)

    so we end up with 2U*dU/dt - U = const

    => dU/dt - const*U^-1 = 1

    This should be solveable
    How do we know that x is independent of t? [If it's not then we certainly can't assume it's constant.]

    I think we lack enough information to be able to find a solution here. Is there more to the question?
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    Thanks so much guys for those helpful suggestions. As Nuodai said it would be better if x were a function of t .

    it may be that the function is quadratic:

    x = ( k - t )^2

    where k is a constant

    then the equation would become:

    dy/dt = y^0.5 - ( k - t )

    & then the tangent field with y and t axes would be appropriate: here the horizontal axis is t: k is 20
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