Author Topic: TuT0701  (Read 453 times)

Yuanxi Gong

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TuT0701
« on: September 27, 2019, 03:16:53 PM »
Question: dy/dx = 2x/(1+2y)  ,  y(2)=0
        ∫(1+2y)dy =∫(2x)dx
              y+y^2 = x^2 + C
when x=2, y=0
                  0+0 = 2^2 + C
C = -4
            y^2 + y = x^2 -4
   y^2 + y + 1/4 = x^2 – 15/4
       (y + 0.5)^2 = x^2 – 15/4
              y + 0.5 = ±√(x^2 – 15/4)
                       y = ±√(x^2 – 15/4) -0.5
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