dy/dx - 2y/(x+1) =(x+1)^(3/2)
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æ±ç¹è§£ï¼è§å¯å¯ä»¥çåºy' = c(x+1)^(5/2)å¯è½æ¯ä¸ªç¹è§£ï¼å¾
人å¾å°
2.5 c (x+1)^(3/2) - 2c (x+1)^(3/2) = 0.5c (x+1)^(3/2)
æ以c=2
ç¹è§£2(x+1)^(5/2)
对äºdy/dx - 2y/(x+1)=0
å¾å°dy/y = 2dx/(x+1)
æ以ln y = 2ln(x+1)+C
æ以æ¹ç¨è§£ä¸º
y=2(x+1)^(5/2) + C(x+1)^2
æx=1ï¼y=1代人å¾å°
1 = 8æ ¹å·2 +4c
c=0.25 - 2æ ¹å·2
y=2(x+1)^(5/2) + C(x+1)^2
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