'-y = √( x2 - y2),y(1) = 0.5
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'-y = √( x2 - y2),y(1) = 0.5
'-y = √( x2 - y2),y(1) = 0.5
'-y = √( x2 - y2),y(1) = 0.5
基本常微分方程类型,解法如下：
令y=ux, x(x du/dx +u)-ux=√(x^2-x^2 u^2 ) ,
可得=|x| du/dx=|x|√(1^2-u^2 )
若x>0, du/√(1^2-u^2 ) =dx/x,
可得arcsinu=lnx+c;
若x0
可得arcsin0.5=ln1+c,
则c=π/6,所以arcsiny/x=lnx+π/6
'-y = √( x2 - y2),y(1) = 0.5
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