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Help with partial fractions C4 Watch

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    I am able to do a, this is needed for part b which i cannot do.
    Here is my answer for part a, in which you need to express [(13-2x)]/[(2x-3)(x+1)] in partial fractions:

    = 4/(2x+3) - 3/(x+1)

    b) Given that y = 4 at x = 2, use your answer to part (a) to find the solution of the differential equation

    dy/dx = [y(13-2x)]/[(2x-3)(x+1)], x > 1.5

    Express your answer in the form y = f(x)

    I am not entirely sure what it means by the solution to the differential equation or how to get to this bit
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    (Original post by sparkling stars)
    I am able to do a, this is needed for part b which i cannot do.
    Here is my answer for part a, in which you need to express [(13-2x)]/[(2x-3)(x+1)] in partial fractions:

    = 4/(2x+3) - 3/(x+1)

    b) Given that y = 4 at x = 2, use your answer to part (a) to find the solution of the differential equation

    dy/dx = [y(13-2x)]/[(2x-3)(x+1)], x > 1.5

    Express your answer in the form y = f(x)

    I am not entirely sure what it means by the solution to the differential equation or how to get to this bit
    Have you solved differential equations by separation of variables before?

    If you have, can you make a start with this? Post your working if you get stuck.
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    (Original post by notnek)
    Have you solved differential equations by separation of variables before?

    If you have, can you make a start with this? Post your working if you get stuck.
    so would it be split into 4/2x-3 - 3/x+1?
    then 4ln(2x-3) - 3ln(x+1)

    what would you do with the y? is it the thing where you differentiate to get dy/dx?
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    (Original post by sparkling stars)
    so would it be split into 4/2x-3 - 3/x+1?
    then 4ln(2x-3) - 3ln(x+1)

    what would you do with the y? is it the thing where you differentiate to get dy/dx?
    Have you studied separation of variables?

    E.g. do you know how to solve \dfrac{dy}{dx}=y^2x ?

    If you don't then you should probably wait until you've been taught this subject.


    Also, the integral of 4/2x-3 is not 4ln(2x-3). Try differentiating 4ln(2x-3) and you'll see it doesn't work.
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    Proper thread here
    Sorry
 
 
 
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