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Second order differential equations!

  1. Oct 26, 2005 #1
    hi guys need some help on diff eqn, ive done the workingout
    and answers but not sure if they are right mind if someone
    can check them for me thanks

    Find the general solution of the differential equation

    dy/dx - 2y = e^(5x)
    i found I(x) = e^ integral (-2 dx) = e^(-2x)
    as I(x) = e^Integral (p(x)) in this case p=-2
    therfore d/dx((e^-2x)*y) = e^(3x) // as we multiply both sides by e^(-2x)

    therefore e^(-2x)y = integral e^(3x)
    e^(-2x)y = e^(3x)/3 + C
    y= e^(3x)/(3*e^(-2x)) + C
    y= 1/3e^(5x) + Ce^(2x)



    2.


    given that z=2xt^3 - cos5x and x^3 + t^2 = 6x
    find dz/dt using a chain rule

    i will use notatio 9 as the "day symbol"
    dz/dt = 9z/9x*dx/dt + dz/dt
    = (2t^3 + 5sin5x)*dx/dt + 6xt^2 !!(3)!!

    then knowing
    x^3 + t^2 = 6x i diff resp to x so
    3x^2*dx/dt - 6dx/dt = -2t
    dx/dt = 2t/(3x^2 - 6)

    i subtituted this back into (3) and got
    = (4t^4 + 10tsin5x + 18x^3t^2 - 36xt^2) / (3x^2-6)


    3. given that z=e^(5x)sin2y and y=x^2 + 5
    find dz/dx as a function of x using a chain rule

    same 9 symbol represtn day

    dz/dx=9z/9x + 9z/9y*dy\dx
    dz/dx = 5e^(5x)sin2y + 2xe^(5x)cos2y

    i subtitued y=x^2 + 5 and got this

    dx/dx = e^(5x) * [5sin(2x^2+10) + 2xcos(2x^2 + 10) ]

    so i hope i did it right

    thanks for the help guys
     
  2. jcsd
  3. Oct 26, 2005 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Yes, that's correct.
    I think that's right but my feeling is you would be better off leaving it as 6xt^2-(4t^4+ 5sin5x)/(3x^2-6)

    You are missing a factor of 2: The derivative of sin 2y wrt y is 2 cos2y and the derivative of x^2+ 5 is 2x: that should be 4x cos(2x^2+ 10).

    Was there a reason for titling this thread "second order differential equations"? There was only one differential equation and it was first order.
     
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