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Math Help - Differential equations know one solution

  1. #1
    Super Member dhiab's Avatar
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    Differential equations know one solution

    1) Solve :

    y' = \cos (x) - y\sin (x) + y^2 <br />
    Know : y = \sin (x) is the solution.

    2 ) Solve :


    y' = 2 + \frac{1}{2}\left( {x - \frac{1}{x}} \right)y - \frac{1}{2}y^2 <br />

    Know :
    y = x + \frac{1}{x} is the solution.
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  2. #2
    Math Engineering Student
    Krizalid's Avatar
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    Ricatti's equations. There's no much here. What have you done?
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  3. #3
    Super Member
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    Quote Originally Posted by dhiab View Post
    1) Solve :

    y' = \cos (x) - y\sin (x) + y^2 <br />
    Know : y = \sin (x) is the solution.
    If we know a particular solution y_p to the Riccati equation:

    y'+Q(x)y+R(x)y^2=P(x)

    then we can obtain the general solution by making the substitution:

    y=\frac{1}{u}+y_p

    and obtain an easier first order equation:

    u'-(2Ry_p+Q)u=R

    Now, when I do that for the first one, I obtain the general solution for this equation:

    y(x)=\sin(x)+\left(c_1 e^{\cos(x)}-e^{\cos(x)}\int e^{-\cos(x)}\right)^{-1}.

    Now, the interesting part I think, is back-substituting that solution into the original Riccati and verifying that it satisfies it. Forgive me if I'm a little lazy and just used Mathematica:

    Code:
    In[61]:=
    y[x_] := Sin[x] + 1/(c1*Exp[Cos[x]] - 
          Exp[Cos[x]]*Integrate[Exp[-Cos[x]], 
            x]); 
    FullSimplify[D[y[x], x] + Sin[x]*y[x] - 
       y[x]^2]
    
    Out[62]= Cos[x]
    I'm sure you could do the second one now.
    Last edited by shawsend; December 17th 2009 at 09:04 AM.
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