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Check My Work for solving this Differential Equation? :)

  1. Feb 3, 2016 #1
    1. The problem statement, all variables and given/known data
    Question: Solve this differential equation
    dy/dt=1+y2 and y(π/4)=-1
    2. Relevant equations

    3. The attempt at a solution
    dy/dt=1+y2 and y(π/4)=-1

    dy/dt=1+y2
    dy=(1+y2)dt
    dy/(1+y2)=dt
    ∫dy(1+y2)=∫dt
    tan-1(y)=t+c
    y(t)=tan(t)+c
    y(π/4)=tan(π/4)+c
    -1=1+c
    -2=c

    answer: y(t)=tan(t)-2
     
  2. jcsd
  3. Feb 3, 2016 #2

    Samy_A

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    Did you check if dy/dt=1+y2 for your solution?

    I think there is an error in this step:
    tan-1(y)=t+c
    y(t)=tan(t)+c
     
  4. Feb 3, 2016 #3
    hmm I'm not sure that I am seeing the error, does it involve the constant being effected by the tan?
     
  5. Feb 3, 2016 #4

    Samy_A

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    Yes.

    There must be an error somewhere, as inspection would show that y(t)=tan(t)-2 doesn't satisfy the differential equation.
     
  6. Feb 3, 2016 #5
    so would it look something like this: y=tan(t+c) ----not my answer but the next step
     
  7. Feb 3, 2016 #6

    Samy_A

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    Yes.
     
  8. Feb 3, 2016 #7
    ok so then then from there using y(π/4)=-1
    the next step would be: -1=tan(π/4+c)
    and then Im guessing the step would be: tan-1(-1)=π/4+c
     
  9. Feb 3, 2016 #8

    Samy_A

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    Yes (no need for guessing, though).
     
  10. Feb 3, 2016 #9
    adding on to the previous steps: -π/4=π/4+c
    -π/2=c

    so the the solution would be:
    y(t)=tan(t-π/2)
     
  11. Feb 3, 2016 #10

    Samy_A

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    Yes.
     
  12. Feb 3, 2016 #11
    Okay thank you very much. You've been very helpful.
     
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