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Math Help - Innocent looking but confusing DE!

  1. #1
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    Innocent looking but confusing DE!

    Hmm am I missing something obvious but how would I solve y"=ay^2?
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  2. #2
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    Multiply both sides by y' and integrate. That'll reduce the order by one. What does that give you?
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  3. #3
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by usagi_killer View Post
    Hmm am I missing something obvious but how would I solve y"=ay^2?
    For this type of second order ODE the 'standard approach' is based on the identity...

    \displaystyle y^{''} = \frac{d y^{'}}{dx}= \frac{d y^{'}}{dy}\ \frac{dy}{dx} = y^{'}\ \frac{d y^{'}}{dy} (1)

    ... so that the equation you have prosed y^{''}= a\ y^{2} becomes...

    \displaystyle y^{'}\ dy^{'} = a\ y^{2}\ dy (2)

    ... that is 'separated variables' and the solution of which is...

     \displaystyle y^{'\ 2}= \frac{2\ a}{3} y^{3} + c_{1} \implies y^{'} = \pm \sqrt{\frac{2\ a}{3}\ y^{3} + c_{1}} (3)

    Also (3) is 'sepatated variables' and its solution is...

    \displaystyle x = \int \frac{d y}{\pm \sqrt{\frac{2\ a}{3}\ y^{3} + c_{1}}} + c_{2} (4)

    ... even if the integral in (4) is not very 'confortable'...



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    \chi \sigma
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