Re: [理工] [工數]-台聯大98
※ 引述《littleyinyo (I AM YIN)》之銘言：
: Solve the initial value problem,xy'=y+√(x^2+y^2),y(2)=0
xdy=ydx+[(x^2+y^2)^0.5]dx
xdy-ydx=[(x^2+y^2)^0.5]dx
x^2d(y/x)=[(x^2+y^2)^0.5]dx
同除以x
xd(y/x)=[(1+(y/x)^2)^0.5]dx
[(1+(y/x)^2)^-0.5]d(y/x)=(1/x)dx
令y/x=sinhz, d(y/x)=conhzdz
dz=(1/x)dx
z=lnx+lnc
(y/x)=sinh(lncx)
代入初值得,c=1/2
代入sinh的定義，化簡得 y= 0.25x^2-1
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