Solve xyy' - x^2√(y^2+1) = (x+1)(y^2+1)

Solve $xyy' - x^2\sqrt{y^2+1} = (x+1)(y^2+1)$.

Explanation of Method

The combination $\\sqrt{y^2+1}$ and $y^2+1$ suggests $u=\\sqrt{y^2+1}$, which turns $yy'$ into $uu'$ and reduces the equation to a first-order linear ODE in $u(x)$.

Substitute $u=\\sqrt{y^2+1}$

Then $u^2=y^2+1$ gives $u\\,du/dx = y\\,dy/dx$, and $y^2=u^2-1$. The original equation $xy\\,y' = x^2\\sqrt{y^2+1}+(x+1)(y^2+1)$ becomes $xu\\,du/dx = x^2u+(x+1)u^2$.

Reduce to a linear equation

Dividing by $u$: $x\\dfrac{du}{dx} - (x+1)u = x^2$, i.e. $\\dfrac{du}{dx} - \\left(1+\\dfrac{1}{x}\\right)u = x$.

Solve with an integrating factor

$\\mu(x) = e^{-\\int(1+1/x)dx} = \\dfrac{e^{-x}}{x}$. Then $\\dfrac{d}{dx}\\!\\left(\\dfrac{ue^{-x}}{x}\\right) = e^{-x}$, so $\\dfrac{ue^{-x}}{x} = -e^{-x}+C$, giving $u = Cxe^x - x$. (Confirmed independently with sympy's dsolve on $xu'-(x+1)u=x^2$, which returns the same family.)

Back-substitute $u=\\sqrt{y^2+1}$

\[ \\sqrt{y^2+1} = Cxe^x - x. \]

Verify

Implicit-differentiating $F=\\sqrt{y^2+1}-(Cxe^x-x)=0$ with sympy and solving for $y'$ gives $y' = \\dfrac{\\sqrt{y^2+1}\\,(Cxe^x+Ce^x-1)}{y}$. Substituting $\\sqrt{y^2+1}=Cxe^x-x$ and $y^2+1=(Cxe^x-x)^2$ into both this expression and into the original equation solved for $y'$, the residual $xyy'-x^2\\sqrt{y^2+1}-(x+1)(y^2+1)$ simplifies symbolically to exactly $0$, confirming the solution.

Result

Final answer:

Boxed result: \[ \\boxed{\\sqrt{y^2+1} = Cxe^x - x} \] equivalently $y^2 = (Cxe^x-x)^2-1$.