4.15.26 \(\sqrt {b^2-y(x)^2} y'(x)=\sqrt {a^2-x^2}\)

ODE
\[ \sqrt {b^2-y(x)^2} y'(x)=\sqrt {a^2-x^2} \] ODE Classification

[_separable]

Book solution method
Separable ODE, Neither variable missing

Mathematica
cpu = 0.0880849 (sec), leaf count = 94

\[\left \{\left \{y(x)\to \text {InverseFunction}\left [\frac {1}{2} \left (\text {$\#$1} \sqrt {b^2-\text {$\#$1}^2}+b^2 \tan ^{-1}\left (\frac {\text {$\#$1}}{\sqrt {b^2-\text {$\#$1}^2}}\right )\right )\& \right ]\left [\frac {1}{2} \left (x \sqrt {a^2-x^2}+a^2 \tan ^{-1}\left (\frac {x}{\sqrt {a^2-x^2}}\right )\right )+c_1\right ]\right \}\right \}\]

Maple
cpu = 0.02 (sec), leaf count = 75

\[ \left \{ {\frac {x}{2}\sqrt {{a}^{2}-{x}^{2}}}+{\frac {{a}^{2}}{2}\arctan \left ({x{\frac {1}{\sqrt {{a}^{2}-{x}^{2}}}}} \right ) }-{\frac {y \relax (x ) }{2}\sqrt {{b}^{2}- \left (y \relax (x ) \right ) ^{2}}}-{\frac {{b}^{2}}{2}\arctan \left ({y \relax (x ) {\frac {1}{\sqrt {{b}^{2}- \left (y \relax (x ) \right ) ^{2}}}}} \right ) }+{\it \_C1}=0 \right \} \] Mathematica raw input

DSolve[Sqrt[b^2 - y[x]^2]*y'[x] == Sqrt[a^2 - x^2],y[x],x]

Mathematica raw output

{{y[x] -> InverseFunction[(b^2*ArcTan[#1/Sqrt[b^2 - #1^2]] + #1*Sqrt[b^2 - #1^2]
)/2 & ][(x*Sqrt[a^2 - x^2] + a^2*ArcTan[x/Sqrt[a^2 - x^2]])/2 + C[1]]}}

Maple raw input

dsolve(diff(y(x),x)*(b^2-y(x)^2)^(1/2) = (a^2-x^2)^(1/2), y(x),'implicit')

Maple raw output

1/2*x*(a^2-x^2)^(1/2)+1/2*a^2*arctan(1/(a^2-x^2)^(1/2)*x)-1/2*y(x)*(b^2-y(x)^2)^
(1/2)-1/2*b^2*arctan(1/(b^2-y(x)^2)^(1/2)*y(x))+_C1 = 0