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Differential response problem

  1. Aug 22, 2014 #1
    I get close to the answer but just not seeing where the integers in the answers are coming from, perhaps forgetting a constant? This is a EE circuit analysis class.

    1. The problem statement, all variables and given/known data

    Given y(t) = dx(t)/dt
    a. Find x(t) if y(t)=
    0 : t<0;​
    e[itex]^{-t}[/itex] : t>0​
    b. Find x(t) for t=1 if x(1)=1 and y(t)=t[itex]^{2}[/itex] for t[itex]\geq[/itex]1

    3. The attempt at a solution

    Part a

    when y(t) = 0:
    ∫0 = ∫[itex]\frac{dx(t)}{dt}[/itex]
    0 = x(t) (this part agrees with given answer)

    when y(t) = e[itex]^{-t}[/itex]:
    ∫ e[itex]^{-t}[/itex]=∫[itex]\frac{dx(t)}{dt}[/itex]

    part b, y(t)=t[itex]^{2}[/itex]:

    ∫t[itex]^{2}[/itex] = ∫[itex]\frac{dx(t)}{dt}[/itex]
    [itex]\frac{1}{3}[/itex]t[itex]^{3}[/itex] = x(t)

    Given answers:

    part a: 0, t[itex]\leq[/itex]0; (1-e[itex]^{-t}[/itex]), t[itex]\geq[/itex]0

    part b: [itex]\frac{1}{3}[/itex](2+t[itex]^{3}[/itex]), t [itex]\geq[/itex]1
  2. jcsd
  3. Aug 22, 2014 #2


    User Avatar

    Staff: Mentor

    Welcome to the PF.

    Yeah, you have to include the constant from the initial condition x(1)=1. Plug t=1 into your solution for b, and plug it into the given solution. Do you see the difference?
  4. Aug 23, 2014 #3
    I do, my answer resolves to 1/3 whereas the correct answer resolves to 1. My calculus is a bit rusty but adding the +C for the constant of integration didn't get me any closer to the 1 in part a or the 2 in part b.
  5. Aug 23, 2014 #4


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    Staff: Mentor

    Could you show your work on adding the constant in part b? The answer given matches the initial condition, and your answer should match it too. The form of their answer looks a little weird, but if you expand terms, it goes into the standard form.
  6. Aug 23, 2014 #5
    I didn't get beyond adding a C to the right side. I think the problem may be with the limits of the integration but I'm not sure, hence my post here.
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