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Math Help - Ordinary Differential Equations!Urgent Help Neede

  1. #1
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    Ordinary Differential Equations!Urgent Help Neede

    Hye all, as a newbie in this forum, i would be really sorry if got any mistake or indirectly breaching the terms and condition in this forum.


    I'm just really2 need a help from anyone of you, please helping me to attain the solution of both questions. The question is on Ordinary Differential Equations topic. This is my important assignment question as it will bring 15% of my career marks (continuous assessment) before final exam. Please really help as that 15% is too important to me..thanks in advanced mates



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    Super Member PaulRS's Avatar
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    1. y'=3+5y thus \frac{y'}{3+5y}=1

    Now integrate: \int_0^{x}\frac{y'}{3+5y}dt=x (let u=y(t) on the LHS)
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    Quote Originally Posted by PaulRS View Post
    1. y'=3+5y thus \frac{y'}{3+5y}=1

    Now integrate: \int_0^{x}\frac{y'}{3+5y}dt=x (let u=y(t) on the LHS)

    thanks..i need to try by my self first and if u dont mind to check my answer after all is it my answer is correct or not
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    Quote Originally Posted by noob View Post
    thanks..i need to try by my self first and if u dont mind to check my answer after all is it my answer is correct or not
    Okay, post your solutions and we'll look at them. However you can check your own work by simply plugging the solution back through the original equation.

    -Dan
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    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by noob View Post
    Hye all, as a newbie in this forum, i would be really sorry if got any mistake or indirectly breaching the terms and condition in this forum.


    I'm just really2 need a help from anyone of you, please helping me to attain the solution of both questions. The question is on Ordinary Differential Equations topic. This is my important assignment question as it will bring 15% of my career marks (continuous assessment) before final exam. Please really help as that 15% is too important to me..thanks in advanced mates



    if my memory is still good, this might help you..

    so for the second one, i remembered we use variation of parameter when we were solving higher orders.

    so first, differentiate (with respect to x): y' + 2xy = x

    we obtain: y'' + 2xy' + 2y = 1

    and start using the method here.
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  6. #6
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    Quote Originally Posted by kalagota View Post
    if my memory is still good, this might help you..

    so for the second one, i remembered we use variation of parameter when we were solving higher orders.

    so first, differentiate (with respect to x): y' + 2xy = x

    we obtain: y'' + 2xy' + 2y = 1

    and start using the method here.

    how can we get y'' + 2xy' + 2y = 1????? is it we get y''+ 2y = 1...where did 2xy' came from??
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  7. #7
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    Quote Originally Posted by topsquark View Post
    Okay, post your solutions and we'll look at them. However you can check your own work by simply plugging the solution back through the original equation.

    -Dan
    hye topsquark, hope u really can help (or the others) this is my answer n im sure i tottally wrong plz check mates..im in my first year of studies and got a bit problem with my maths

    answer for first questions


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    Quote Originally Posted by noob View Post
    how can we get y'' + 2xy' + 2y = 1????? is it we get y''+ 2y = 1...where did 2xy' came from??
    Product rule: (2xy)' = 2x'y + 2xy' = 2y+2xy'

    So lets differentiate:

    y' + 2xy - x = 0

    (y' + 2xy - x)' = 0

    (y')' +(2xy)' - (x)' = 0

    y'' + 2y + 2xy' - 1 = 0

    y'' + 2xy' +2y = 1
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  9. #9
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    Quote Originally Posted by Isomorphism View Post
    Product rule: (2xy)' = 2x'y + 2xy' = 2y+2xy'

    So lets differentiate:

    y' + 2xy - x = 0

    (y' + 2xy - x)' = 0

    (y')' +(2xy)' - (x)' = 0

    y'' + 2y + 2xy' - 1 = 0

    y'' + 2xy' +2y = 1
    thanks mates..its really help
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  10. #10
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    Quote Originally Posted by noob View Post
    hye topsquark, hope u really can help (or the others) this is my answer n im sure i tottally wrong plz check mates..im in my first year of studies and got a bit problem with my maths

    answer for first questions



    how bout my first solutions..please comment and add up some ideas if it wrong thanks very2 much
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  11. #11
    Lord of certain Rings
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    Quote Originally Posted by noob View Post
    how bout my first solutions..please comment and add up some ideas if it wrong thanks very2 much
    I cant read it,why dont you type it out in a post?
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    Quote Originally Posted by noob View Post
    hye topsquark, hope u really can help (or the others) this is my answer n im sure i tottally wrong plz check mates..im in my first year of studies and got a bit problem with my maths

    answer for first questions


    Quote Originally Posted by Isomorphism View Post
    I cant read it,why dont you type it out in a post?
    this is my answer and surely it's wrong because im also did not really sure what am i doing, please refer to my first post regarding the question by the way thanks for your willingness





    p/s: really like your siggy..so meaningful
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  13. #13
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    Its correct

    I cant read the last two lines but it must be:

    5y = e^{5x+5C} - 3

    y = \frac15 e^{5x+5C} - \frac35

    However I would add a point that e^{5C} is a constant, so call it A.Thus:

    y = \frac{A}5 e^{5x} - \frac35
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  14. #14
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    Quote Originally Posted by Isomorphism View Post
    Its correct

    I cant read the last two lines but it must be:

    5y = e^{5x+5C} - 3

    y = \frac15 e^{5x+5C} - \frac35

    However I would add a point that e^{5C} is a constant, so call it A.Thus:

    y = \frac{A}5 e^{5x} - \frac35

    thank you very much, from the question needed, is it i can left the answer just like this??one more things that quite confusing me is when did we need to consider the integral factor?? some question in book using an integral factor method
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  15. #15
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    Quote Originally Posted by noob View Post
    thank you very much, from the question needed, is it i can left the answer just like this??
    Yes you can.

    one more things that quite confusing me is when did we need to consider the integral factor?? some question in book using an integral factor method
    You mean "integrating factor"? There is no need for it here, since the differential equation is in separable form..
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