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MAT244--2019F
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TuT0701
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Topic: TuT0701 (Read 884 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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TuT0701