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Math Help - Is my solution correct?

  1. #1
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    Is my solution correct?

    can someone please check my solution?
    Attached Thumbnails Attached Thumbnails Is my solution correct?-img_1262.jpg   Is my solution correct?-gash.jpg  
    Last edited by CaptainBlack; April 16th 2010 at 03:54 AM.
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  2. #2
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    Quote Originally Posted by noteiler View Post
    can someone please check my solution?
    Dear noteiler,

    Can you please tell me how you obtained the equation shown below,

    C_{1}-\frac{1}{p}=\frac{y}{p^{2}}
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  3. #3
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    Quote Originally Posted by Sudharaka View Post
    Dear noteiler,

    Can you please tell me how you obtained the equation shown below,

    C_{1}-\frac{1}{p}=\frac{y}{p^{2}}
    by integrating both sides of
    \frac{1}{p^2}=\frac{dy}{p^2} - \frac{2dpy}{p^3}
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  4. #4
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    Quote Originally Posted by noteiler View Post
    by integrating both sides of
    \frac{1}{p^2}=\frac{dy}{p^2} - \frac{2dpy}{p^3}
    Dear noteiler,

    You can't obtain, C_{1}-\frac{1}{p}=\frac{y}{p^{2}} by integrating both sides of \frac{dp}{p^2}=\frac{dy}{p^2}-\frac{2y}{p^3}. This is because,

    \frac{dp}{p^2}=\frac{dy}{p^2}-\frac{2y}{p^3}dp

    \int{\frac{dp}{p^2}}=\color{red}{\int{\frac{dy}{p^  2}}}-\color{red}{\int{\frac{2y}{p^3}dp}}.

    The highlighted integrations could not be evaluated beacause they contain two variables whose relation we don't know.

    But instead try,

    (p+2y)\frac{dp}{dy}=p

    Dividing by y, \left(\frac{p}{y}+2\right)\frac{dp}{dy}=\frac{p}{y  }

    Now use the substitution, u=\frac{p}{y}

    Hope you could countinue from here.
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  5. #5
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    Quote Originally Posted by Sudharaka View Post
    Dear noteiler,

    You can't obtain, C_{1}-\frac{1}{p}=\frac{y}{p^{2}} by integrating both sides of \frac{dp}{p^2}=\frac{dy}{p^2}-\frac{2y}{p^3}. This is because,

    \frac{dp}{p^2}=\frac{dy}{p^2}-\frac{2y}{p^3}dp

    \int{\frac{dp}{p^2}}=\color{red}{\int{\frac{dy}{p^  2}}}-\color{red}{\int{\frac{2y}{p^3}dp}}.

    The highlighted integrations could not be evaluated beacause they contain two variables whose relation we don't know.
    it doesn't work that way, i don't integrate two seperate functions, but i integrate the whole 2-argument function. i mean,
    d(\frac{y}{p^2})=\frac{dy}{p^2}-\frac{2ydp}{p^3}
    left side is the complete differential or whatever it is called in English, so we can integrate it. of course, there is a propability that i am wrong or making silly mistake, but i use this very often and so far i has been correct
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  6. #6
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    Quote Originally Posted by noteiler View Post
    it doesn't work that way, i don't integrate two seperate functions, but i integrate the whole 2-argument function. i mean,
    d(\frac{y}{p^2})=\frac{dy}{p^2}-\frac{2ydp}{p^3}
    left side is the complete differential or whatever it is called in English, so we can integrate it. of course, there is a propability that i am wrong or making silly mistake, but i use this very often and so far i has been correct
    You're OK with what you've done except at the very end. THe solution I got was

    x = 2 c_1 p - \ln p + c_2
    y = c_1p^2 - p.
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  7. #7
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    Quote Originally Posted by Danny View Post
    You're OK with what you've done except at the very end. THe solution I got was

    x = 2 c_1 p - \ln p + c_2
    y = c_1p^2 - p.
    thanks for confirming, i've already started to worry
    but isn't that wright
    \ln p + c_2 = \ln p + \ln c_2 = \ln c_2p
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  8. #8
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    Quote Originally Posted by noteiler View Post
    thanks for confirming, i've already started to worry
    but isn't that wright
    \ln p + c_2 = \ln p + \ln c_2 = \ln c_2p
    You're right. Point well taken.
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  9. #9
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    Quote Originally Posted by noteiler View Post
    it doesn't work that way, i don't integrate two seperate functions, but i integrate the whole 2-argument function. i mean,
    d(\frac{y}{p^2})=\frac{dy}{p^2}-\frac{2ydp}{p^3}
    left side is the complete differential or whatever it is called in English, so we can integrate it. of course, there is a propability that i am wrong or making silly mistake, but i use this very often and so far i has been correct
    Dear noteiler,

    Thank you very much for this post. I learnt something new because of your thread. Although your answer is almost correct you are missing out a important modulus sign.

    x=2c_{1}p-lnc_{2}{\color{red}\mid{p}\mid}

    y=c_{1}p^{2}-p

    Hope this will help you.
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