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Cauchy prob.

  1. May 3, 2010 #1
    [tex]y'=\sin (x+y+3)[/tex]
    [tex] y(0)=-3 [/tex]

    I tried substituting [tex] x+y+3=u [/tex] and solving I get
    [tex] \tan (u(x)) - \sec (u(x)) = x [/tex]

    but what the heck can I do now?
  2. jcsd
  3. May 4, 2010 #2
    You get
    [tex]\frac{d u}{dx} = 1+\sin(u)[/tex]

    and then

    [tex]\int\frac{d u}{1+\sin(u)} = x+C,[/tex]
    where [tex]C[/tex] is an arbitrary constant, or

    [tex]-\frac{2}{\tan[\frac{u}{2}]+1}= x+C[/tex]

    so the general solution to your ODE is

    [tex]y(x) = -2\arctan(\frac{2+x+C}{x+C})-x-3[/tex]

    Substituting [tex]x=0[/tex] you find that [tex]C=-2[/tex], so particular solution with condition [tex]y(0)=-3[/tex]

    [tex]y(x) = -2\arctan(\frac{x}{x-2})-x-3[/tex]
  4. May 4, 2010 #3
    Thanks, it seems fine now.
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