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2nd Order differential help!! watch

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    d^2y/dx^2 + dy/dx = e^-x

    I'm having some trouble with this, think i must be missing something!

    First i found the complementary function:

    m^2 + m = 0
    m(m+1) = 0
    so: m=0 or m=-1

    Therefore complementary function is: y= Ae^0x + Be^-x

    Which is: y=A + Be^-x

    That part is correct! However i am stuck at finding the particular integral.
    Let y=ke^-x
    dy/dx= -ke^-x
    d^2y/dx^2= ke^-x

    Substituting into the original equation gives:
    ke^-x - ke^-x = e^-x
    Therefore k-k = 1
    0=1!!!!!
    Have i made a mistake or is there a special trick needed for this one?!
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    (Original post by jmzcherry)
    d^2y/dx^2 + dy/dx = e^-x

    I'm having some trouble with this, think i must be missing something!

    First i found the complementary function:

    m^2 + m = 0
    m(m+1) = 0
    so: m=0 or m=-1

    Therefore complementary function is: y= Ae^0x + Be^-x

    Which is: y=A + Be^-x

    That part is correct! However i am stuck at finding the particular integral.
    Let y=ke^-x
    dy/dx= -ke^-x
    d^2y/dx^2= ke^-x

    Substituting into the original equation gives:
    ke^-x - ke^-x = e^-x
    Therefore k-k = 1
    0=1!!!!!
    Have i made a mistake or is there a special trick needed for this one?!
    You already have ke^-x in your particular integral, so that can't work.
    Try kxe^-x.
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    ok thanks
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    If your particular integral had been (A+Bx)e^-x then you would have needed to try k(x^2)(e^-x) and so on.
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    if i have e^-2x (Acosx + Bsinx)

    and f(x) is sin2x, can i ley y=asinx + bcosx for my particular integral???

    Or do i have to use asin^2 x + bcos^2 x???
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    (Original post by jmzcherry)
    if i have e^-2x (Acosx + Bsinx)

    and f(x) is sin2x, can i ley y=asinx + bcosx for my particular integral???

    Or do i have to use asin^2 x + bcos^2 x???
    y=asin2x+bcos2x would be your particular integral, I'm not sure why you want to square them.
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    If a situation arose where i had y= Acosx + Bsinx as the complementary function and f(x) =cosx, what would i use as my particular integral seeing as cosx is already included in the complementary function?
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    (Original post by jmzcherry)
    If a situation arose where i had y= Acosx + Bsinx as the complementary function and f(x) =cosx, what would i use as my particular integral seeing as cosx is already included in the complementary function?
    I haven't seen that happen before but i'd try Axcosx+Bxsinx, as you usually just multiply by an x.
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    Thats what i thought! Just want to think of every possibility incase anything nasty crops up in the exam!
 
 
 
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