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A.m Q10

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  • 03-06-2007 4:01am
    #1
    sd123
    Closed Accounts Posts: 534 ✭✭✭


    ok, 4 in the morning now, just wondering does anyone know how to do 2nd order differential eqtns cos i havnt a clue! looking through the papers, it hasn't come up since 83 at least. will it come up this year by ANY chance. also, whats the chances of power coming up as 10 (b)?? has only come up twice since 83.:o


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    #2
    ninebeanrows
    Registered Users Posts: 12,778 ✭✭✭✭ninebeanrows


    You got up at 4 o clock!! I struggled to get up now!!

    Your going to be wrecked if you keep that up ;)


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    #3
    Assez Bien
    Closed Accounts Posts: 74 ✭✭Assez Bien


    Oh my God i didn't go to bed til 5 dis mornin.....n it wasn't studyin i was at......these forums are beginning 2 seriously depress me as i have finally realised what an utter waster i am! Sorry this is of no benefit 2 d question u asked!!


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    #4
    alan4cult
    Registered Users Posts: 1,583 ✭✭✭alan4cult


    Solve d2y/dx2 = 3 dy/dx given that y=0 and dy/dx=1 when x=0

    d2y/dx2 = 3 dy/dx

    Let v = dy/dx

    then dv/dx = d2y/dx2

    dv/dx = 3v (Changing Original Equation)

    dv = 3v dx

    dv/3v = dx

    integral (1/3v)dv = integral (1)dx
    1/3 integral (1/v)dv = x
    1/3(ln v) + c = x

    x = 0 dy/dx=v=1

    1/3ln (1)+c = 0
    c=0

    therefore 1/3(ln v) = x
    lnv = 3x
    e^3x = v

    v = dy/dx = e^3x

    dy = e^3x(dx)

    integral 1(dy) = integral e^3x(dx)
    y = [e^3x]/3 + K

    y = 0 when x = 0

    0 = [e^0]/3 + K
    0 = 1/3 + K
    K = -1/3

    therefore y = [e^3x]/3 - 1/3

    y= 1/3(e^3x - 1)

    Just put in your own variable, in this case v, then use calculus to get it out and then to get an equation with y and x in it.


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