1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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:
    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
    I integrate both sides to get
    C is the constant, I combine the constants from both sides to one value.
    Multiplying both sides by the negative,
    Now i e both sides
    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


    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]
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook