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    3).

    A curve passes through (0,1) and satisfies the differential equation dy/dx= (4-y^2)^1/2. Find its equation and sketch its graph.

    Basically I have no idea how to integrate it?

    Any hints would be appreciated.
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    (Original post by Super199)
    3).

    A curve passes through (0,1) and satisfies the differential equation dy/dx= (4-y^2)^1/2. Find its equation and sketch its graph.

    Basically I have no idea how to integrate it?

    Any hints would be appreciated.
    use the substitution y = 2sinθ
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    (Original post by Super199)
    3).

    A curve passes through (0,1) and satisfies the differential equation dy/dx= (4-y^2)^1/2. Find its equation and sketch its graph.

    Basically I have no idea how to integrate it?

    Any hints would be appreciated.
    Use a substitution, y=2\sin(\theta) or y=2\cos(\theta) if you prefer. or look it up in a table - it's a standard result.
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    (Original post by TeeEm)
    use the substitution y = 2sinθ
    (Original post by joostan)
    Use a substitution, y=2\sin(\theta) or y=2\cos(\theta) if you prefer. or look it up in a table - it's a standard result.
    I think I am missing something here. What is the standard result I am supposed to see? I am not sure if were taught this, or it may be some c3/c4 stuff I have forgotten.
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    (Original post by Super199)
    I think I am missing something here. What is the standard result I am supposed to see? I am not sure if were taught this, or it may be some c3/c4 stuff I have forgotten.
    The integral:
    \displaystyle\int \dfrac{1}{\sqrt{a^2-x^2}} \ dx
    for a given constant a is a standard result.
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    (Original post by joostan)
    The integral:
    \displaystyle\int \dfrac{1}{\sqrt{a^2-x^2}} \ dx
    for a given constant a is a standard result.
    oh right yeah that gives inverse sine right. sin^-1(x/a)+c
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    (Original post by Super199)
    I think I am missing something here. What is the standard result I am supposed to see? I am not sure if were taught this, or it may be some c3/c4 stuff I have forgotten.
    It is only standard for further mathematicians,,, Just use the substitution
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    (Original post by TeeEm)
    It is only standard for further mathematicians,,, Just use the substitution
    It seems like a lot more effort. How does knowing the standard result help?
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    (Original post by Super199)
    It seems like a lot more effort.
    I am sorry to hear that.

    (Original post by Super199)
    How does knowing the standard result help?
    you can have the answer in 3 lines
    (assuming you have gone the long hard substitution way a few times to have an appreciation of the techniques involved)
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    (Original post by TeeEm)
    I am sorry to hear that.



    you can have the answer in 3 lines
    (assuming you have gone the long hard substitution way a few times to have an appreciation of the techniques involved)
    Decent, care to inform me on how that is done.

    Like I probably could have done it if it was 1/(4-y^2). As that is sin^-1(y/2) +c . But yeah we have one over that so how does the integral change?
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    (Original post by Super199)
    Decent, care to inform me on how that is done.

    Like I probably could have done it if it was 1/(4-y^2). As that is sin^-1(y/2) +c . But yeah we have one over that so how does the integral change?
    I am not following

    once you separate variables you do have 1/over
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    (Original post by TeeEm)
    I am not following

    once you separate variables you do have 1/over
    ah I got it now cheers bae
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    (Original post by Super199)
    ah I got it now cheers bae
    all the best
 
 
 
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