※ 引述《asdf322505 ()》之銘言：
: 2√xy-y
: y'=------------
: x
: 1
: ANS;√y =√x( 1 - ----)
: cx
: 誰能交交我該怎麼算
: 我算答案怪怪的
原式： y' = 2[x^(-1/2)]y^(1/2) - x^(-1)y
y' + x^(-1)y = 2[x^(-1/2)]y^(1/2) ...(★)→ Bernoulli's form
Let u = y^[1-(1/2)] = y^(1/2) , that is, y = u^2 , dy/du = 2u
by Chain rule dy/dx = (dy/du) * (du/dx) = 2u(du/dx) substuting into (★)
2u(du/dx) + x^(-1)u^2 = 2[x^(-1/2)]u
Dividing by 2u：(du/dx) + (1/2)x^(-1)u = [x^(-1/2)] →Linear ODE
Integrator factor I(x) = exp[∫(1/2)x^(-1)dx] = x^(1/2)
After multiplying integrator factor I(x)
d
---[x^(1/2)u] = 1 , x^(1/2)u= x + c
dx
u = y^(1/2) = x(1/2) + cx(-1/2)
∴ y^(1/2) = x(1/2) [1-cx^(-1)]
有錯還煩請高手不吝於指正 <(_ _)>
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