求通解 y''＝〔(2y-1)/(y²+1)〕(y')² 要过程谢谢
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该微分方程属于缺 x 型，即缺自变量型。
设 y' = p , 则 y'' = dp/dx = (dp/dy)(dy/dx) = pdp/dy
微分方程化为 pdp/dy = [(2y-1)/(y^2+1)]p^2
p = 0 , y = C;
dp/p = ](2y-1)/(y^2+1)]dy ,
lnp = ln(1+y^2) - arctany + lnC1
p = dy/dx = C1(1+y^2)/e^arctany
e^arctanydy/(1+y^2) = C1dx
e^arctanydarctany = C1dx
e^arctany = C1x + C2
设 y' = p , 则 y'' = dp/dx = (dp/dy)(dy/dx) = pdp/dy
微分方程化为 pdp/dy = [(2y-1)/(y^2+1)]p^2
p = 0 , y = C;
dp/p = ](2y-1)/(y^2+1)]dy ,
lnp = ln(1+y^2) - arctany + lnC1
p = dy/dx = C1(1+y^2)/e^arctany
e^arctanydy/(1+y^2) = C1dx
e^arctanydarctany = C1dx
e^arctany = C1x + C2
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