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Math Help - please help with thease DEs

  1. #1
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    please help with thease DEs

    could you please help me to solve thease 4 DEs :
    1. i've integrated the last function with maple but the result was very complicated and an answer in my book was simple, but i can't see it.
    2.how can i get the other solutions? in the answers there was mentioned something with ln.
    3. what i am doing wrong, because answer in the book is something else, like sin and sh functions.
    4.what should i do next?
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  2. #2
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    Quote Originally Posted by noteiler View Post
    could you please help me to solve thease 4 DEs :
    1. i've integrated the last function with maple but the result was very complicated and an answer in my book was simple, but i can't see it.
    2.how can i get the other solutions? in the answers there was mentioned something with ln.
    3. what i am doing wrong, because answer in the book is something else, like sin and sh functions.
    4.what should i do next?
    I will answer your first question:

    \sqrt{C_1 - \frac{1}{y^2}} = \frac{\sqrt{C_1 y^2 - 1}}{y} (assuming y > 0). So the substitution u = C_1 y^2 - 1 is clearly suggested.
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  3. #3
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    just hit me how to solve 4.
    C1y=y'/(1-y')
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  4. #4
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    For the second...

    \frac{d^2y}{dx^2} = 2y\,\frac{dy}{dx}

    \int{\frac{d^2y}{dx^2}\,dx} = \int{2y\,\frac{dy}{dx}\,dx}

    \int{\frac{d^2y}{dx^2}\,dx} = \int{2y\,dy}

    \frac{dy}{dx} = y^2 + C.


    Now this is a separable DE

    \frac{1}{y^2 + C}\,\frac{dy}{dx} = 1

    \int{\frac{1}{y^2 + C}\,\frac{dy}{dx}\,dx} = \int{1\,dx}

    \int{\frac{1}{y^2 + C}\,dy} = \int{1\,dx}.


    The RHS is easy to evaluate. The LHS requires trig substitution.

    Let y = \sqrt{C}\tan{\theta} so that dy = \sqrt{C}\sec^2{\theta}\,d\theta.

    Note that \theta = \arctan{\frac{y}{\sqrt{C}}}

    LHS becomes

    \int{\frac{1}{y^2 + C}\,dy} = \int{\frac{1}{(\sqrt{C}\tan{\theta})^2 + C}\,\sqrt{C}\sec^2{\theta}\,d\theta}

     = \int{\frac{\sqrt{C}\sec^2{\theta}}{C\tan^2{\theta} + C}\,d\theta}

     = \int{\frac{\sqrt{C}\sec^2{\theta}}{C(\tan^2{\theta  } + 1)}\,d\theta}

     = \int{\frac{\sqrt{C}\sec^2{\theta}}{C\sec^2{\theta}  }\,d\theta}

     = \int{\frac{\sqrt{C}}{C}\,d\theta}

     = \frac{\sqrt{C}}{C}\theta

     = \frac{\sqrt{C}}{C}\arctan{\frac{y}{\sqrt{C}}}.


    So that means...

    \int{\frac{1}{y^2 + C}\,dy} = \int{1\,dx}

    \frac{\sqrt{C}}{C}\arctan{\frac{y}{\sqrt{C}}} = x + D

    \arctan{\frac{y}{\sqrt{C}}} = \sqrt{C}(x + D)

    \frac{y}{\sqrt{C}} = \tan{[\sqrt{C}(x + D)]}

    y = \sqrt{C}\tan{[\sqrt{C}(x + D)]}.


    Now you can use your boundary conditions to solve for C and D.
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  5. #5
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    Quote Originally Posted by noteiler View Post
    just hit me how to solve 4.
    C1y=y'/(1-y')
    Note that \frac{p}{1-p} = -1 + \frac{1}{1-p}.
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