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Homework Help: Solve First order linear differential equation, initial conditions

  1. Jan 13, 2010 #1
    1. The problem statement, all variables and given/known data
    The problem is given as follows:
    Solve
    dy/dt + y = 0.5, y(t=0)=1


    2. Relevant equations



    3. The attempt at a solution
    I separate the y terms from the t terms, which gives me
    dy(-y+0.5)=dt
    I integrate both sides to get
    -ln(-y+0.5)=t+C
    C is the constant, I combine the constants from both sides to one value.
    Multiplying both sides by the negative,
    ln(-y+0.5)=-t-C
    Now i e both sides
    -y+0.5=e^(-t-C)
    Therefore I can simplify to y=e^(-t-C)+0.5, which is my solution
    Since y(0)=1,
    e^(-C) = 0.5
    I don't know what exactly I am supposed to do with that...
    Was my answer correct?
    Please advise, Thank you!
     
    Last edited: Jan 13, 2010
  2. jcsd
  3. Jan 13, 2010 #2
    surely you mean [itex]e^{-C}=1.5[/itex]
     
  4. Jan 13, 2010 #3
    Wrote the signs wrong, fixed now.
     
  5. Jan 13, 2010 #4

    HallsofIvy

    User Avatar
    Science Advisor

    dy/(-y+0.5)= dt

    [itex]-y+0.5= e^{-t-C}[/itex] is NOT equivalent to [itex]y= e^{-t-C}+0.5[/itex].

    Subtracting 0.5 from each side of [itex]-y+0.5= e^{-t-C}[/itex] gives [itex]-y= e^{-t-C}- 0.5[/itex] and then, multiplying both sides by -1, [itex]y= 0.5- e^{-t-C}[/itex].

    Now, taking y= 1 when t= 0, [itex]1= 0.5- e^{-C}[/itex] so [itex]e^{-C}= 0.5[/itex] still.
    And [itex]e^{-t-C}= e^{-t}e^{-C}[/itex]. Since [itex]e^{-C}= 0.5[/itex], [itex]e^{-t-C}= 0.5e^{-t}[/itex] and your solution is [itex]y= 0.5- 0.5e^{-t}[/itex]
     
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