math:calculus

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math:calculus [2019/12/22 23:07] – phreazermath:calculus [2019/12/22 23:10] – [Differential equations] phreazer
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 $-(1/y)  = 1/2 x^2 + c_2 - c_1$ $(c_2 - c_1 = c)$ $-(1/y)  = 1/2 x^2 + c_2 - c_1$ $(c_2 - c_1 = c)$
  
-$y = \frac{-1}{(1/2 x^2 + c)}$+$y = \frac{-1}{(1/2 x^2 + c)}$ (y is a function not value, obviously)
  • math/calculus.txt
  • Last modified: 2019/12/22 23:28
  • by phreazer