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Does this look correct? (DE)

  1. Sep 26, 2013 #1
    1. The problem statement, all variables and given/known data
    [tex](x^{2}+1)(tan y)(y')=x[/tex]


    2. Relevant equations



    3. The attempt at a solution
    [tex](x^{2}+1)(tan y)(\frac{dy}{dx})=x[/tex]
    [tex](tan y)dy=\frac{x}{x^{2}+1}dx[/tex]
    [tex]\int tanydy=\int \frac{x}{x^2+1}dx \\ -ln|cos y|=\frac{ln|x^{2}+1|}{2}+C[/tex]
    [tex]e^{-ln(cosy)}=e^{\frac{ln(x^{2}+1)}{2}+C}[/tex]
    [tex]\frac{1}{cosy}=e^{C}(\sqrt{x^2+1})[/tex]
    [tex]cosy=\frac{1}{D\sqrt{x^{2}+1}} \implies y=cos^{-1}\bigg(\frac{1}{D\sqrt{x^2+1}}\bigg)[/tex]
    D represents e^C just fyi.

    Wondering if this is correct and if my work makes sense.
     
    Last edited: Sep 26, 2013
  2. jcsd
  3. Sep 26, 2013 #2

    SteamKing

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    You can always differentiate your y expression and see if it satisfies the original DE.
     
  4. Sep 26, 2013 #3
    I don't even know how to differentiate that LOL
     
  5. Sep 27, 2013 #4

    HallsofIvy

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    Strange! You are solving differential equations but don't know how to differentiate? The derivative of arc-cosine(x) is [itex]1/\sqrt{1- x^2}[/itex] and the derivative of [itex](x^2+ 1)^{1/2}[/itex] is [itex](1/2)(x^2+ 1)^{-1/2}(2x)= x(x^2+ 1)^{-1/2}= x/\sqrt{x^2+ 1}[/itex].

    Putting those together with the chain rule, the derivative of [itex]cos^{-1}(1/D\sqrt{x^2+ 1})[/itex] is
    [tex]\frac{1}{\sqrt{1- \frac{1}{D^2}\frac{1}{x^2+ 1}}}\frac{x}{D\sqrt{x^2+ 1}}= \frac{x}{\sqrt{D^2(x^2+ 1)- 1}}[/tex]
     
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