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Finding a solution?

  1. Sep 25, 2004 #1
    Finding a solution??

    Find all solutions to

    y'= 3y + 15

    and solve the initial value problem-
    y(0)= -1

    so would it be y= 3/2x^2 + 15y + C?????

    if not please share what would it be!!!!
  2. jcsd
  3. Sep 25, 2004 #2
    This should be in Calculus and Analysis. (sorry, I'm not sure about your quesiton.)
    Last edited: Sep 25, 2004
  4. Sep 25, 2004 #3
    This problem is easily solved by seperation of variables (which I see you did). However, I'm pretty sure you meant 3/2y^2. if y(0) = -1, what does that tell you about the value of C?
  5. Sep 25, 2004 #4
    Ofcourse! It's been too long...
  6. Sep 25, 2004 #5
    This is a first order linear differential equation. It will be instructive to write this equation as [tex]\frac{dy}{dx} = 3y + 15[/tex]. Because the right side contains [tex]y(x)[/tex] you can't directly integrate the equation as it looks like you tried to do. Instead you can write it so all the [tex]y[/tex] stuff is on the left and all the [tex]x[/tex] is on the right: [tex]\frac{dy}{3y + 15} = dx[/tex]. NOW we can integrate both sides to get [tex]\int_{-1}^{y}\frac{dy}{3y + 15} = \int_{0}^{x}dx[/tex]. Notice that I've already included the initial condition that [tex]y = -1[/tex] when [tex]x=0[/tex] in my limits of integration. It is now a simple matter of integrating the left side and then solving for [tex]y(x)[/tex]. The answer I got was [tex]y(x) = 4e^{3x}-5[/tex].
    Last edited: Sep 26, 2004
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