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Please check my math on diff equation with initial conditions question (easy for you)

  1. Nov 24, 2009 #1
    1. The problem statement, all variables and given/known data

    Solve the differential equation: y' + 2y = te^-2t with initial conditions y(1) = 0?
    Please show me what im doing wrong... (or right).. also please dont show me shortcut i need to know where im going wrong thanks.



    2. Relevant equations



    3. The attempt at a solution

    first i multiplied whole equation by integrating factor e^2t (e^integral(2) = e^2t)
    after rearranging this gave me

    e^2t(y) = integ te^-2t * e^2t

    i then simplified RHS to get:
    e^2t(y) = integ t

    then i integrated RHS to get

    e^2t(y) = (t^2)/2 + c

    then i isolated y to get:

    y = ((t^2)e^-2t) / 2 + ce^-2t

    then i applied the initial condition y(1) = 0
    this gave me

    0 = e^-2 / 2 + ce^-2

    then i multiplied equation by 2

    0 = e^-2 + ce^-2

    then i factored out e^-2

    0 = e^-2 (1 + c)

    then i reasoned that c must = 1 and put this back into the explicit equation

    y = ((t^2)e^-2t) /2 + (1) e^-2t

    is this correct? the book answer says y = (t^2 -1) e^-2t /2

    THANKS FOR ANY HELP
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Nov 24, 2009 #2

    diazona

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    Homework Helper

    Re: Please check my math on diff equation with initial conditions question (easy for

    Look for a couple of arithmetic mistakes after you used the initial condition y(1)=0.
     
  4. Nov 24, 2009 #3

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Re: Please check my math on diff equation with initial conditions question (easy for

    Then you reasoned wrong. c is not equal to 1.

     
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