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    Hi

    haven't done these for a while so I might be missing something simple but can anyone give me any pointers on...

    dy/dx = (c^2 - (k^2)(y^2))^0.5

    thanks
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    PS Helper
    This is a separable differential equation; that is, it's in the form \dfrac{dy}{dx} = f(x)g(y) (where f,g are functions of x,y, respectively). In general, the method is to rearrange the equation to get \dfrac{1}{g(y)} \dfrac{dy}{dx} = f(x). Integrating both sides w.r.t. x then gives \displaystyle \int \dfrac{1}{g(y)}\, \text{d}y = \int f(x)\, \text{d}x.

    If you can't work out what f(x) and g(x) are

    No really, try to find them first

    You can use f(x) = 1 and g(x) = \text{the stuff on the RHS}.
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    (Original post by nuodai)
    This is a separable differential equation; that is, it's in the form \dfrac{dy}{dx} = f(x)g(y) (where f,g are functions of x,y, respectively). In general, the method is to rearrange the equation to get \dfrac{1}{g(y)} \dfrac{dy}{dx} = f(x). Integrating both sides w.r.t. x then gives \displaystyle \int \dfrac{1}{g(y)}\, \text{d}y = \int f(x)\, \text{d}x.

    If you can't work out what f(x) and g(x) are

    No really, try to find them first

    You can use f(x) = 1 and g(x) = \text{the stuff on the RHS}.

    Thanks
    I did actually separate them like that! But I thought it was wrong because I didn't know how to integrate 1/RHS wrt y...
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    (Original post by bonbon)
    Thanks
    I did actually separate them like that! But I thought it was wrong because I didn't know how to integrate 1/RHS wrt y...
    Ah yes, I should have mentioned: the substitution y = \frac{c}{k} \sin \theta might help. Or if you're a formula book addict, look for the integral in the form \displaystyle \int \dfrac{1}{\sqrt{a^2-b^2x^2}}\, \text{d}x.
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    (Original post by nuodai)
    Ah yes, I should have mentioned: the substitution y = \frac{c}{k} \sin \theta might help. Or if you're a formula book addict, look for the integral in the form \displaystyle \int \dfrac{1}{\sqrt{a^2-b^2x^2}}\, \text{d}x.

    ah ok thanks. should get it from here thanks
 
 
 
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