Question

y'' = √1-(y')^2
срочно решить!!!пожалуйста

Answer 1

$y''=sqrt{1-(y')^2}\\y'=z(y); ,; ; y''=z'(y)cdot y'=z'cdot z\\z'z=sqrt{1-z^2}\\frac{dz}{dy}=frac{sqrt{1-z^2}}{z}; ; ,; ; int frac{z, dz}{sqrt{1-z^2}}=int dy\\-frac{1}{2}cdot 2sqrt{1-z^2}=y+C_1; ,; ; sqrt{1-z^2}=-(y+C_1)\\1-z^2=(y+C_1)^2; ; ,; ; z^2=1-(y+C_1)^2\\(y')^2=1-(y+C_1)^2; ; ,; ; y'=pm sqrt{1-(y+C_1)^2}\\frac{dy}{dx}=pm sqrt{1-(y+C_1)^2}; ; ,; ; int frac{dy}{sqrt{1-(y+C_1)^2}}=pm int dx\\arcsin(y+C_1)=pm x+C_2\\y+C_1=sin(C_2pm x)\\y=sin(C_2pm x)-C_1$