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    • Thread Starter

    Yet another question from me! Sorry, have just finished a 80-question review exercise but there are a few I'm stuck on! Can anyone help me with the following:

    Find the solution of the differential equation:
    dy/dx = y/(1+x)(3+x)
    given that y=2 at x=1. Express your answer in the form y^2=f(x)

    In the earlier part of the question I worked out 1/(1+x)(3+x) in partial fractions to be 1/2(1+x) - 1/2(3+x). So I assumed that to solve that differential eqtn i used:

    Int 1/y dy= Int x. dx

    And continued to solve it but I don't get the right answer of y^2 = 8(1+x)/(3+x)


    The solution is given by:

    INT 1/y dy = INT 1/(1+x)(3+x) dx

    You are right about the partial fractions, so the solution is then

    INT 1/y dy = (1/2) INT 1/(1+x) - 1/(3+x) dx
    ln|y| = (1/2)[ ln|1+x| - ln|3+x| + lnA ] (write the constant as lnA rather than C)

    So, mutiplying by 2 and combing the logarithms on the right gives

    2ln|y| = ln| A(1+x)/(3+x) |

    Note that 2ln|y| = ln|y^2|, therefore

    ln|y^2| = ln| A(1+x)/(3+x) |
    y^2 = A(1+x)/(3+x)

    Sub y=2,x=1:

    4 = A(2/4) => A=8

    Which gives the result

    y^2 = 8(1+x)/(3+x)

    I hope this helps.
    • Thread Starter

    Thanks! btw, not sure why I called it vectors as the title!!!!! What was I thinking!

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