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Solving a differential equation

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    • Thread Starter

    If someone could explain part (c) to me that would be great!
    I could swear the 2 on the top of the fraction just disappears : /
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    Separating the variables gives

     \displaystyle \int \frac{3}{y}dy=\int \frac{2}{x^2-1} dx .

    So using part a you get

     \displaystyle \int \frac{3}{y}dy = \int \left ( \frac{1}{x-1}-\frac{1}{x+1} \right ) dx .

    I split it this way so that the RHS is identical to the expression that you split into partial fractions in part A.

    You can also use the partial fractions you have worked out in the previous question.
    • Thread Starter

    Ah just realised the two hasn't just disappeared I was meant to replace the fraction with the partial fraction form, which is you already integrate in part (b), and the 'y' at the top switches places with the 'dx' at the bottom of 'dy/dx' as shown in the mark scheme, which also means the left hand side become 'ln(y)' rather than 'y' however this is easily sorted out later. The rest is just simplifying and re-arranging to find '+C'.

    Thanks for the help guys!
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Updated: April 11, 2016
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