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 = 0.370905 (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 = 0.023 (sec), leaf count = 75

$\left [\frac {x \sqrt {a^{2}-x^{2}}}{2}+\frac {a^{2} \arctan \left (\frac {x}{\sqrt {a^{2}-x^{2}}}\right )}{2}-\frac {y \left (x \right ) \sqrt {b^{2}-y \left (x \right )^{2}}}{2}-\frac {b^{2} \arctan \left (\frac {y \left (x \right )}{\sqrt {b^{2}-y \left (x \right )^{2}}}\right )}{2}+\textit {\_C1} = 0\right ]$ 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))

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]