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Second Order ODE using Reduction of Order watch

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    Separable?
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    (Original post by DFranklin)
    Separable?
    It goes to \int\frac{1}{1+\frac{c}{y}} dy=\int dx

    But how to integrate LHS? Log?
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    Firstly, that's not what it comes to.
    Secondly, neither integral (the one you've given, or the one it should be) is difficult to integrate.

    You are not going to be able to solve these problems reliably if you don't understand how to do basic (A-level) integration.
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    (Original post by DFranklin)
    Firstly, that's not what it comes to.
    Secondly, neither integral (the one you've given, or the one it should be) is difficult to integrate.

    You are not going to be able to solve these problems reliably if you don't understand how to do basic (A-level) integration.
    Sorry it should be a y in place of the 1 in the denominator.

    All I learnt at A level was that \int\frac{1}{1+x}dx=ln|1+x|+c and \int\frac{1}{1-x}dx=-ln|1-x|+c

    Is it one of those where the top is close to the derivative of the bottom? So \frac{1}{2}ln|y^2+c| ?
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    If you're not sure if you've got the integral right, differentiate it to check.
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    (Original post by DFranklin)
    If you're not sure if you've got the integral right, differentiate it to check.
    I get y=\pm\sqrt{Ae^{2x}-c}

    Does that seem about right?
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    I have no idea.

    It is straightforward to differentiate and substitute into the original equation to see if it works. If you are not sure if your solution is correct, this is what you should do. (You might find it helps to calculate d/dx (y^2) as well).
 
 
 
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