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Is this equation solvable and if so, how?

  1. Feb 23, 2010 #1
    xy'=3y+(cosx)x^4

    I don't understand how to solve it since it's non-separable.
     
  2. jcsd
  3. Feb 23, 2010 #2
    y'/x^3-3y/x^4=cosx

    d/dx(y/x^3)=cosx

    y/x^3=sinx

    y=(sinx)x^3
     
  4. Feb 23, 2010 #3
    There's more to Differential Equations than just the separable ones. A LOT more.
     
  5. Feb 24, 2010 #4
    Surely

    d/dx(y/x^3)=cosx

    y/x^3=sinx+C

    y=(sinx+C)x^3

    where C is an arbitrary constant.
     
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