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    Can the above be solved using the following formula:



    Sin x differentiates to Cos x but I am unsure of how to treat the '3' in front of the y.

    I know you can divide by Sin x and use the alternative method (integrating factor method) but I wanted to know if its still possible using the 'exact' derivative method..
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    (Original post by GPODT)


    Can the above be solved using the following formula:



    Sin x differentiates to Cos x but I am unsure of how to treat the '3' in front of the y.

    I know you can divide by Sin x and use the alternative method (integrating factor method) but I wanted to know if its still possible using the 'exact' derivative method..
    As it stands, you do not have an 'exact' derivative. The purpose of the integrating factor is precisely to put the equation in a form where it is exact
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    (Original post by davros)
    As it stands, you do not have an 'exact' derivative. The purpose of the integrating factor is precisely to put the equation in a form where it is exact
    Sorry if I was unclear but my question is, is it possible to use this:



    To solve the question? I mean instead of f'(x)y (as in the formula) we have f'(x)3y..
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    (Original post by GPODT)
    Sorry if I was unclear but my question is, is it possible to use this:



    To solve the question? I mean instead of f'(x)y (as in the formula) we have f'(x)3y..
    I must be missing something here!

    f(x)y' + f'(x)y = (fy)' is an exact differential, which is what you get when you use an integrating factor.

    You can't choose y to be your dependent function in one bit of the equation and 3y in another bit - it won't work.
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    (Original post by davros)
    I must be missing something here!

    f(x)y' + f'(x)y = (fy)' is an exact differential, which is what you get when you use an integrating factor.

    You can't choose y to be your dependent function in one bit of the equation and 3y in another bit - it won't work.
    (Original post by GPODT)
    x
    I think this is the solution OP is looking for?

    \sin x \frac{dy}{dx} + 3y \cos x = \mathrm{cosec} x

    Can be rearranged to:

    \frac{dy}{dx} + 3(\cot x )y = \mathrm{cosec}^2 x

    Hence, your integrating factor is e^\int3\cot x\ dx

    Or is that not the answer you were looking for?
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    (Original post by Anythingoo1)
    I think this is the solution OP is looking for?

    \sin x \frac{dy}{dx} + 3y \cos x = \mathrm{cosec} x

    Can be rearranged to:

    \frac{dy}{dx} + 3(\cot x )y = \mathrm{cosec}^2 x

    Hence, your integrating factor is e^ \int 3 \cot x\ dx

    Or is that not the answer you were looking for?
    Beat me to it, and that's correct

    Slightly tricky Q compared to the usual 1st order ODE Qs
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    (Original post by Indeterminate)
    Beat me to it, and that's correct

    Slightly tricky Q compared to the usual 1st order ODE Qs
    Aha, you'll get the next one

    Btw, I'm having a problem writing e^ int... basically the integrating factor in latex, as you can see, it doesn't want me to give the integrating bit as a power, do you know what I'm supposed to do?
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    (Original post by Anythingoo1)
    Aha, you'll get the next one

    Btw, I'm having a problem writing e^ int... basically the integrating factor in latex, as you can see, it doesn't want me to give the integrating bit as a power, do you know what I'm supposed to do?
    \displaystyle e^{ \int \cot x \ dx}

    in latex:

    \displaystyle e^{\int \cot x \ dx}
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    (Original post by Anythingoo1)
    I think this is the solution OP is looking for?

    \sin x \frac{dy}{dx} + 3y \cos x = \mathrm{cosec} x

    Can be rearranged to:

    \frac{dy}{dx} + 3(\cot x )y = \mathrm{cosec}^2 x

    Hence, your integrating factor is e^\int3\cot x\ dx

    Or is that not the answer you were looking for?
    That would make sense! I was confused by what the OP was trying to say in his first post
 
 
 
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