Question

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

Answer 1

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