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Math Help - help me with differential equation

  1. #1
    Newbie Black Kawairothlite's Avatar
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    help me with differential equation

    hey all this is my first post and im sorry that i begin asking bt well ill try to help too whenever i can...

    this is about second grade differential equations...

    consider the equation

    xy''-(x+N)y'+Ny=0 where N>=0

    show that y_1=e^x is a solution and the second solution can be in the form

    y_2(x)=e^x \int x^n e^{-x} dx
    Last edited by Black Kawairothlite; October 26th 2008 at 07:51 PM.
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  2. #2
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    Quote Originally Posted by Black Kawairothlite View Post
    hey all this is my first post and im sorry that i begin asking bt well ill try to help too whenever i can...

    this is about second grade differential equations...

    consider the equation xy''-(x+N)y'+Ny=0 where N>=0

    show that y1= e^x is a solution and the second solution can be in the form

    y2(x)=e^x \int x^n e^-x dx
    I assume you can calculate y_1' and y_1'' from y_1 = e^x. Substitute the results into the left hand side of the differential equation and show that the result simplifies to zero.

    Try finding a solution of the form y_2(x) = y_1(x) u(x) = e^x u(x). Calculate y_2' and y_2'' and substitute the results into the left hand side of the differential equation. Simplify and solve the resulting differential equation for u(x).
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  3. #3
    Newbie Black Kawairothlite's Avatar
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    thanks for your answer Mr. Fantastic...ill show ya what i did cuz i have a mistake and i don't know where it is...so let's see:


    first of all i took the first solution and i evaluated it in the equation and yes it was a solution (i had already done it before but i didn't do it well)

    so then i did grade reduction to get the second solution which in the exercise said it had to be:

    y_2= e^x\int x^Ne^{-x}dx

    well so this is what i got:

    y_2= e^xv(x) where v(x) is my unknown solution

    y_2'=e^xv(x)+e^xv'(x)

    y_2'=e^xv(x)+2e^xv'(x)+e^xv''(x)

    then replacing on the diff eq. (which was xy''-(x+N)y'+Ny=0 with N>0)

    x(e^xv(x)+2e^xv'(x)+e^xv''(x))-(x+N)(e^xv(x)+e^xv'(x))+N(e^xv(x))=0

    some sums and products later...

    v''xe^x-v'e^x(x+N)=0

    then i suppouse that u=v'(x) so the new equation is:

    u'xe^x-ue^x(x+N)=0 which is a separable diff. equation

    here is the problem i think...

    u'=u((x+N)/x)

    du/dx=u((x+N)/x)

    du/u=((x+N)/x)dx

    (while i was writing this i noticed my mistake, i was not ok because i put du/u=dx/((x+N)/x)) let's c if i'm right...)

    ln u=x+ln x^N

    u=e^xx^N

    remember that u=v' so:

    v=\int e^xx^Ndx and finally:

    y_2=e^x\int e^xx^Ndx but i`m still missing the minus where do i get it (omg it worked pretty good)

    and the other point i care is that i gotta prove that this second solution is a polynomial of N grade...

    Last edited by Black Kawairothlite; October 26th 2008 at 07:53 PM.
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  4. #4
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    Quote Originally Posted by Black Kawairothlite View Post
    thanks for your answer Mr. Fantastic...ill show ya what i did cuz i have a mistake and i don't know where it is...so let's see:


    first of all i took the first solution and i evaluated it in the equation and yes it was a solution (i had already done it before but i didn't do it well)

    so then i did grade reduction to get the second solution which in the exercise said it had to be:

    y_2= e^x\int x^Ne^{-x}dx

    well so this is what i got:

    y_2= e^xv(x) where v(x) is my unknown solution

    y_2'=e^xv(x)+e^xv'(x)

    y_2'=e^xv(x)+2e^xv'(x)+e^xv''(x)

    then replacing on the diff eq. (which was xy''-(x+N)y'+Ny=0 with N>0)

    x(e^xv(x)+2e^xv'(x)+e^xv''(x))-(x+N)(e^xv(x)+e^xv'(x))+N(e^xv(x))=0

    some sums and products later...

    v''xe^x-v'e^x(x+N)=0 Mr F says: I think you'll find this should be {\color{red}v'' x e^x + v' e^x (x-N) = 0}. And you can divide out the common factor of e^x.

    then i suppouse that u=v'(x) so the new equation is:

    u'xe^x-ue^x(x+N)=0 which is a separable diff. equation

    here is the problem i think...

    u'=u((x+N)/x)

    du/dx=u((x+N)/x)

    du/u=((x+N)/x)dx

    (while i was writing this i noticed my mistake, i was not ok because i put du/u=dx/((x+N)/x)) let's c if i'm right...)

    ln u=x+ln x^N

    u=e^xx^N

    remember that u=v' so:

    v=\int e^xx^Ndx and finally:

    y_2=e^x\int e^xx^Ndx but i`m still missing the minus where do i get it (omg it worked pretty good)

    and the other point i care is that i gotta prove that this second solution is a polynomial of N grade...

    Good job showing all your work
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  5. #5
    Newbie Black Kawairothlite's Avatar
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    thank you M. Fantastic i'm learning LaTex code here lol and btw i already did what u said you can c that when i put this:

    u'=u((x+N)/x)
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  6. #6
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    Quote Originally Posted by Black Kawairothlite View Post
    thank you M. Fantastic i'm learning LaTex code here lol and btw i already did what u said you can c that when i put this:

    u'=u((x+N)/x)
    Not so. From what I said it follows that

    \frac{du}{dx} = u \, \frac{(N - x)}{x}.
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  7. #7
    Newbie Black Kawairothlite's Avatar
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    Wonderful i got the answer!! thank you Mr. Fantastic you're fantastic lol btw what do u think about the other thing? ( i mean about proving that the second solution is a polynomial of N grade)

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  8. #8
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    Quote Originally Posted by Black Kawairothlite View Post
    Wonderful i got the answer!! thank you Mr. Fantastic you're fantastic lol btw what do u think about the other thing? ( i mean about proving that the second solution is a polynomial of N grade)

    I'd suggest evaluating the integral (use integration by parts N times - try substituting some concrete values for N first to get the feel of things. N= 1, 2, 3). Then multiply the result by e^x.
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  9. #9
    Newbie Black Kawairothlite's Avatar
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    well just for the record im gonna show the second part of the problem i hope this could help someone in the future and if i'm wrong correct me plz...


    well with the changes that MR. Fantastic adviced me i got the answer of the solution which was:

    y_2(x)=e^x \int x^n e^{-x} dx

    and as he said i did the integration by parts N times:

    u=x^N

    du= Nx^{N-1}dx

    dv=e^{-x}dx

    v=-e^{-x}

    then:

    \int e^{-x}x^Ndx= -x^Ne^{-x} + \int e^{-x}Nx^{N-1}dx

    again and again: (i gonna do it directly)

    \int e^{-x}x^Ndx= -x^Ne^{-x} - Nx^{N-1}e^{-x} + \int e^{-x}(N^2-N)x^{N-2}dx

    \int e^{-x}x^Ndx= -x^Ne^{-x} - (N^2-N)x^{N-2}e^{-x} + \int e^{-x}x^{N-3}(N^3-3N^2-2N)dx+ ... - \frac{{d^M(-x^N)}}{{dx^M}}e^{-x} +\int \frac{{d^{M+1}(-x^N)}}{{dx^{M+1}}}e^{-x}dx

    where M,N>=0

    then the second solution...

    y_2= e^x(-x^Ne^{-x} - (N^2-N)x^{N-2}e^{-x} + \int e^{-x}x^{N-3}(N^3-3N^2-2N)dx+ ... - \frac{{d^M(-x^N)}}{{dx^M}}e^{-x} +\int \frac{{d^{M+1}(-x^N)}}{{dx^{M+1}}}e^{-x}dx)

    and finally

    y_2=-x^N - (N^2-N)x^{N-2} + e^x\int e^{-x}x^{N-3}(N^3-3N^2-2N)dx+...- \frac{{d^M(-x^N)}}{{dx^M}} +e^x\int \frac{{d^{M+1}(-x^N)}}{{dx^{M+1}}}e^{-x}dx

    as you can see is a polynomial of N grade

    (yesterday)
    i've checked it again and found some errors but right now i'm too tired to solve them so i hope you help me with that

    (today)
    ok i think it's done
    Last edited by Black Kawairothlite; October 28th 2008 at 09:11 AM. Reason: check, done
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