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Finding a solution to this?

  1. Sep 25, 2004 #1
    Find all solutions to

    y'= 3y + 15

    and solve the initial value problem-
    y'=3y+15
    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

    mathwonk

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    just separate variables. i.e. dy/dx = 3y + 15, so

    dy = (3y+15)dx
    so

    dy/(y+15) = dx. now integrate both sides. and keep the constant of integration on one side.
     
  4. Sep 25, 2004 #3

    arildno

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    Hint:
    Rewrite your differential equation as:
    y'=3(y+5)
    Introduce the new function v(x)=y(x)+5, which implies:
    v'(x)=y'(x), and your differential equation in y now reads in v:
    v'=3v
    Can you take solve it on your own from here?
     
  5. Sep 26, 2004 #4
    can someone please solve it entirely, i dont get it at all.
     
  6. Sep 26, 2004 #5

    arildno

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    Since
    v'=3v, clearly
    v(x)=Ke^{3x} for some K.
    Hence, y(x)=Ke^{3x}-5
    K can be determined by the initial condition.
     
  7. Sep 26, 2004 #6
    [tex] dy/dx = 3y+15 [/tex]
    [tex] 1/(3y+15) dy = dx [/tex]
    [tex]\int dy/(3y+15) = \int dx [/tex]

    u = 3y + 15 1/3 du = dy

    [tex] 1/3 \int du /(u)= x + c_1 [/tex]
    [tex] 1/3(ln(u)+c_2) = x + c_1 [/tex]
    [tex] 1/3 (ln (3y + 15) + c_2) = x + c_1 [/tex]
    [tex] ln (3y + 15) = 3(x + c_3) [/tex]
    [tex] y + 5 = e^{x + c} [/tex]

    [tex] y = (e^{x + c}) - 5 [/tex]

    [tex] -1 = e^{-1 + c} - 5 [/tex]
    [tex] -1+5 = 4 = e^{-1}e^{c}[/tex]
    [tex]4e = e^c[/tex]
    [tex]ln(4e) = c [/tex]
    [tex] 2 ln(2) + 1 = c [/tex]

    [tex]y = e^{x + 2 ln 2 + 1} - 5 [/tex]
     
    Last edited: Sep 26, 2004
  8. Sep 26, 2004 #7

    arildno

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    Phymath:
    You've forgotten to exponentiate your "3" in front of (x+c3)..
     
  9. Sep 26, 2004 #8
    the x value is 0 guys, not -1, but thanks for the help, i appreciate it.

    I have another problem that i tried solving in the dy/dx way, but it gets a bit too complex.

    Find the solutions of the differential equation

    y'-4y=-10sin(2x)

    which has the form y= Asin(2x)+Bcos(2x)
     
  10. Sep 26, 2004 #9

    arildno

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    Just plug in for y' and y; you'll get 2 equations by which you may determine A and B
     
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