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1st order differential equation

  1. Feb 21, 2014 #1
    Find the general solution of the first order differential equation [itex](y+x^{2}y)\frac{dy}{dx}=3x+xy^{2}[/itex], with [itex]y(1)=1[/itex].



    My attempt:
    [tex]\frac{y}{3+y^{2}}dy=\frac{x}{1+x^{2}}dx ∴ \frac{1}{2}\int \frac{2y}{3+y^{2}}dy=\frac{1}{2}\int \frac{2x}{1+x^2}dx[/tex]
    [tex]=\frac{1}{2}ln|3+y^{2}|=\frac{1}{2}ln|1+x^{2}|+C[/tex] ∴ [tex]y^{2}+3=x^{2}+1+e^{C}[/tex] ∴ [tex]y=\pm\sqrt{x^{2}-2+e^{C}}[/tex] Applying boundary conditions gives [tex]1=\pm\sqrt{1^{2}-2+e^{C}} \Rightarrow e^{C}=2[/tex] Therefore [tex]y=\pm x[/tex].


    Is this right?
     
  2. jcsd
  3. Feb 21, 2014 #2

    Mentallic

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    Homework Helper

    Not quite right here. Pay particular attention to your exponential rules. Remember that

    [tex]a^{b+c}=a^ba^c[/tex]
     
  4. Feb 21, 2014 #3
    Ohhh! Of course...silly me.
     
  5. Feb 21, 2014 #4

    Mentallic

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    Don't worry, I answered a whole homework problem set on this topic with the exact same mistake in each and every question haha :biggrin:
     
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