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Thread: Differential Equations Help

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    Aura's Avatar
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    Differential Equations Help

    Can anyone pls help with this math question, all steps would be appreciated:

    (dy/dx)=(y/x-3), where x can't equal 3. Find the particular solution to y=f(x) with the initial condition f(-1)=1
    The answer is y=(-1/4)x+(3/4)


    pls help, will luv u
    muchas gracias mes amis



    please @ me if you can help, I need to study so I won't be able to check frequently
    Last edited by Aura; 06-01-2017 at 08:32 PM.
    i love luna

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    Aura's Avatar
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    i love luna

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    teddy's Avatar
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    good luck! i can't math at all :c

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    Aura (06-01-2017)

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    @(you need an account to see links)
    It's a separable differential equation.
    Just came back from work and kind of tired but:
    (you need an account to see links)
    (you need an account to see links)
    Just keep in mind how to do the constant part though I think those websites will show you.

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    Aura (06-01-2017)

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    Quote Originally Posted by Aura View Post
    Can anyone pls help with this math question, all steps would be appreciated:

    (dy/dx)=(y/x-3), where x can't equal 3. Find the particular solution to y=f(x) with the initial condition f(-1)=1
    The answer is y=(-1/4)x+(3/4)


    pls help, will luv u
    muchas gracias mes amis



    please @ me if you can help, I need to study so I won't be able to check frequently
    Okay, its a separable equation.

    First, you know x can't equal 3 because the denominator can't be 0. So if x-3=0, x=3, therefore x=/=3

    So, you multiply both sides by dx and divide both sides by y to get

    dy/y=dx/(x-3)

    Then you integrate both sides.

    integration of 1/y is ln|y|+C
    integration of 1/(x-3) is ln|x-3|+C

    So you get

    ln|y|=ln|x-3|+C1

    using the property of logarithms
    ln|y|=ln|e^C1(x-3)|

    multiply by e both sides

    y=e^c1(x-3) (since e^c1 is constant)
    y=c(x-3)

    sub in initial conditions y=1, x=-1

    1=c((-1)-3)
    1=c(-4)
    -1/4=c

    plug in

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

    therefore your final equation is y=-1/4x+3/4 where x=/=3
    Last edited by kat.; 06-02-2017 at 07:40 AM.

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    Aura (06-02-2017)

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    Quote Originally Posted by ckatg View Post
    Okay, its a separable equation.

    First, you know x can't equal 3 because the denominator can't be 0. So if x-3=0, x=3, therefore x=/=3

    So, you multiply both sides by dx and divide both sides by y to get

    dy/y=dx/(x-3)

    Then you integrate both sides.

    integration of 1/y is ln|y|+C
    integration of 1/(x-3) is ln|x-3|+C

    So you get

    ln|y|=ln|x-3|+C1

    using the property of logarithms
    ln|y|=ln|e^C1(x-3)|

    multiply by e both sides

    y=e^c1(x-3) (since e^c1 is constant)
    y=c(x-3)

    sub in initial conditions y=1, x=-1

    1=c((-1)-3)
    1=c(-4)
    -1/4=c

    plug in

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

    therefore your final equation is y=-1/4x+3/4 where x=/=3
    I got to the c=-1/4 part, I don't know why it was so confusing afterwards haha. Thanks so much!
    i love luna

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