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Math Help - Tricky integration, possibly Integration Factor

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    Super Member craig's Avatar
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    Tricky integration, possibly Integration Factor

    Hi, for this problem I would normally use the Integration factor, but as it's y^2 I'm not too sure what to do.

    x\frac{dy}{dx} - y^2 = 4

    Have divided through by x, as I would normally in a question like this:

    \frac{dy}{dx} - \frac{y^2}{x} = \frac{4}{x}.

    Not too sure where to go from here however...

    Thanks in advance for the help
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    Moo
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    Hello,
    Quote Originally Posted by craig View Post
    Hi, for this problem I would normally use the Integration factor, but as it's y^2 I'm not too sure what to do.

    x\frac{dy}{dx} - y^2 = 4

    Have divided through by x, as I would normally in a question like this:

    \frac{dy}{dx} - \frac{y^2}{x} = \frac{4}{x}.

    Not too sure where to go from here however...

    Thanks in advance for the help
    Here is how you can do

    x\frac{dy}{dx}=4+y^2

    \Rightarrow \frac{dy}{4+y^2}=\frac{dx}{x}

    Now, can you deal with that ?
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    Super Member craig's Avatar
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    Quote Originally Posted by Moo View Post
    Hello,

    Here is how you can do

    x\frac{dy}{dx}=4+y^2

    \Rightarrow \frac{dy}{4+y^2}=\frac{dx}{x}

    Now, can you deal with that ?
    Thanks for the reply I'm afraid I'm not familiar with that notation though.

    Is that the same as writing

    \int \frac{1}{4+y^2} = \int \frac{1}{x} ?

    Sorry for the lack of understanding
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    Moo
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    Quote Originally Posted by craig View Post
    Thanks for the reply I'm afraid I'm not familiar with that notation though.

    Is that the same as writing

    \int \frac{1}{4+y^2} = \int \frac{1}{x} ?

    Sorry for the lack of understanding
    Yes, but you have to write the dy and dx !

    From the equality I wrote, you can integrate from 0 to t :

    \int_0^t \frac{1}{4+y^2} ~dy=\int_0^t \frac 1x ~dx

    This is called a separable ODE (see here : http://www.usna.edu/MathDept/CDP/DE/...Separation.pdf )
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    Super Member craig's Avatar
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    Quote Originally Posted by Moo View Post
    Yes, but you have to write the dy and dx !

    From the equality I wrote, you can integrate from 0 to t :

    \int_0^t \frac{1}{|4+y^2|} ~dy=\int_0^t \frac 1x ~dx

    This is called a separable ODE (see here : http://www.usna.edu/MathDept/CDP/DE/...Separation.pdf )
    Thank you for this

    So for the indefinite integral \int \frac{1}{4+y^2} ~dy=\int \frac 1x ~dx

    Would you get?

    \frac{1}{2y} \ln{(y^2 + 4)} = \ln(x) + C ?

    Thanks again
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    Quote Originally Posted by craig View Post
    Thank you for this

    So for the indefinite integral \int \frac{1}{4+y^2} ~dy=\int \frac 1x ~dx

    Would you get?

    \frac{1}{2y} \ln{(y^2 + 4)} = \ln(x) + C ?

    Thanks again
    No \int\frac{dy}{4+y^2} = \frac{1}{2} \tan^{-1}\frac{y}{2} + c
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    Super Member craig's Avatar
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    *hits head on desk*

    Of course it is, it's even in my formula book!

    Thank for pointing that out, I think I'll wake myself up before attempting any more of these
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