##### 4.15.25 $$\sqrt {b^2+y(x)^2} y'(x)=\sqrt {a^2+x^2}$$

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 = 0.408987 (sec), leaf count = 89

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

Maple
cpu = 0.018 (sec), leaf count = 67

$\left [\frac {x \sqrt {a^{2}+x^{2}}}{2}+\frac {a^{2} \ln \left (x +\sqrt {a^{2}+x^{2}}\right )}{2}-\frac {y \left (x \right ) \sqrt {b^{2}+y \left (x \right )^{2}}}{2}-\frac {b^{2} \ln \left (y \left (x \right )+\sqrt {b^{2}+y \left (x \right )^{2}}\right )}{2}+\textit {\_C1} = 0\right ]$ 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*Log[#1 + Sqrt[b^2 + #1^2]])/2 + (#1*Sqrt[b^2 + #1
^2])/2 & ][(x*Sqrt[a^2 + x^2])/2 + C[1] + (a^2*Log[x + Sqrt[a^2 + x^2]])/2]}}

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