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Differential equations - 2nd order nonhomogenous eq'n

  1. Mar 23, 2007 #1
    differential equations - 2nd order homogenous eq'n

    sorry the title should read 2nd order homogenous eq'n, not nonhomogenous

    Find the general solution of the equation:
    (1+t^2)d^2y/dt^2 - 2t dy/dt + 2y = 0, given that y1(t) = t is one solution.

    My attempt:
    divided equation by 1+t^2
    d^2y/dt^2 - 2t/1+t^2 dy/dt + 2/1+t^2 y = 0

    using u(t) eq'n given:
    u(t) = exp(-integ(- 2t/t+t^2 dt)) / y1^2(t)
    let x = 1+t^2
    dx/dt = 2t
    dt = dx/2t
    u(t) = exp(integ(2t/x dx/2t)) / t^2
    u(t) = exp(integ(1/x dx)) / t^2
    u(t) = exp(ln x) / t^2
    u(t) = e^ln(1+t^2) / t^2
    u(t) = e^ln(1+t^2) / t^2
    u(t) = (1+t^2) / t^2
    u(t) = (1+t^2)/(t^2)

    using y2(t) eq'n given:
    y2(t) = t integ (u(t)dt)
    y2(t) = t integ ((1+t^2)/(t^2)dt)
    y2(t) = t [ integ(1/t^2 dt) + integ (1 dt)]
    y2(t) = t[-1/t + t]
    y2(t) = -1 + t^2

    y(t) = c1y1 + c2y2
    y(t) = c1t + c2[-1 + t^2]
    y(t) = c1t - c2[t^2 - 1]

    can someone please confirm whether i am doing this correct?

    Last edited: Mar 23, 2007
  2. jcsd
  3. Mar 23, 2007 #2


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    Staff Emeritus
    Science Advisor

    What "u(t) eq'n" are you talking about? And why write "y1^2(t)" when you know y1= t?

    and what "y2(t) eq'n" were you given?
    Are you using formulas in your textbook? We can't very well tell whether what you are doing is correct or not if we don't know what formulas you are using.

    Since you did not explain what you are doing I can't tell whether it is or is not correct. I can say, since it is obvious, that your final y satisfies the differential equation. Surely you had already seen that.
  4. Mar 23, 2007 #3
    the equations i used are the following:
    u(t) = exp(-integ(b/a dt))
    y2(t) = y1(t) integ (u(t) dt)
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