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About simple differential equations

  1. Mar 18, 2012 #1
    when i solve dy/dt= y-b

    (1/y-b)(dy/dt)=1
    d(ln│y-b│)/dt=1

    when i integrate both sides respect to t,
    ln│y-b│=t+c (c is a constant)
    y=±e^(at+c)+b
    =±c1*e^at + b (c1 is a constant)

    then the book replaces ±c1 with c2 (constant)
    but isn't it wrong to do so? Because c2 can't show that it can have two answers.
     
  2. jcsd
  3. Mar 18, 2012 #2

    HallsofIvy

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    There aren't "two solutions". There is a single "general solution" which, for some given "initial value", gives a single solution. [itex]e^{at+c}= e^c e^{at}[/itex]. Since c is an arbitrary constant, so is [itex]e^c[/itex] except that it must be positive since e to any power is positive- your "[itex]c_1[/itex] must be a positive number. Putting "[itex]\pm[/itex]" gets you negative values but is still not quite enough since [itex]\pm ce^{at}[/itex] still cannot be 0. Yet, y= b for all t certainly is a solution so your [itex]y(t)= \pm c_1e^{at}+b[/itex] is not complete- it misses y identically 0. That can be fixed by replacing "[itex]\pm c_1[/itex]" with [itex]c_2[/itex] which can be a postive, negative, or 0.
     
  4. Mar 18, 2012 #3
    So do you mean that c1 means a negative number or positive number or 0 and identiying only one of them? By the way i do not get why c1 can be 0. since c1= e^c1 >0, c1 can't be 0.
     
    Last edited: Mar 18, 2012
  5. Mar 18, 2012 #4

    HallsofIvy

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    I mean that c1 can be any single real number, negative, positive, or 0. Which number depends on what additional information (such as an initial value) you are given.

    Yes, that is exactly what is wrong with your solution. You can't have c1 equal to 0 so you cannot get y= b. But y= b for all x clearly is a solution.
     
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