Solve the following differential equation explicitly. Use lower case c\displaystyle {c}c for the constant of integration.
y′1−x2+1−y2=0\displaystyle {y}'\sqrt{{{1}-{x}^{{2}}}}+\sqrt{{{1}-{y}^{{2}}}}={0}y′1−x2+1−y2=0
Hint: Use the identities ∫11−u2du=sin−1(u)+c\displaystyle \int\frac{{1}}{\sqrt{{{1}-{u}^{{2}}}}}{d}{u}={{\sin}^{{-{{1}}}}{\left({u}\right)}}+{c}∫1−u21du=sin−1(u)+c, −∫11−u2du=cos−1(u)+c\displaystyle -\int\frac{{1}}{\sqrt{{{1}-{u}^{{2}}}}}{d}{u}={{\cos}^{{-{{1}}}}{\left({u}\right)}}+{c}−∫1−u21du=cos−1(u)+c and sin(A+B)=sin(A)cos(B)+sin(B)cos(A)\displaystyle {\sin{{\left({A}+{B}\right)}}}={\sin{{\left({A}\right)}}}{\cos{{\left({B}\right)}}}+{\sin{{\left({B}\right)}}}{\cos{{\left({A}\right)}}}sin(A+B)=sin(A)cos(B)+sin(B)cos(A).
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