Solve the differential equation x[(y')^2 + e^(2y)] = -2y'

Solve the differential equation \[ x\big[(y')^2 + e^{2y}\big] = -2y'. \]

1. Reduce to a quadratic in \(p=y'\)

\(xp^2+2p+xe^{2y}=0\) gives \[ y' = \frac{-1\pm\sqrt{1-x^2e^{2y}}}{x}.\qquad(3) \] Case 2 (minus sign) is squared back and checked to reproduce \(xp^2+2p+xe^{2y}=0\) exactly — this squaring-back step is legitimate here because it is used only to confirm the branch already selected by the quadratic formula, not to introduce a new branch, so no extraneous solutions are smuggled in.

2. Substitute \(u=e^y\), then \(v=xu\)

With \(u=e^y\), \(y'=u'/u\), equation (2) becomes (after multiplying by \(u^2\)) \[ x(u')^2+2uu'+xu^4=0 \;\Rightarrow\; u' = \frac{u(-1\pm\sqrt{1-x^2u^2})}{x}. \] Taking the "\(-\)" branch and setting \(v=xu\) (so \(v'=u+xu'\)) gives, after simplification, \[ \frac{dv}{dx} = -\frac{v}{x}\sqrt{1-v^2}. \qquad\Longrightarrow\qquad \frac{dv}{v\sqrt{1-v^2}} = -\frac{dx}{x}. \]

3. Verify the antiderivative

The claimed antiderivative \(\displaystyle\int\frac{dv}{v\sqrt{1-v^2}} = -\ln\left|\frac{\sqrt{1-v^2}+1}{v}\right|+C_1\) was checked by differentiating the right-hand side in Sympy: the derivative simplifies to exactly \(\dfrac{1}{v\sqrt{1-v^2}}\), confirming the antiderivative is correct as written (no sign error here).

4. Integrate and back-substitute

Integrating and solving, \(\dfrac{\sqrt{1-v^2}+1}{v}=Kx\), and with \(v=xu=xe^y\) this becomes \[ \sqrt{1-x^2e^{2y}}+1 = Kx^2e^y. \] This implicit relation was verified directly against the original quadratic equation \(x[(y')^2+e^{2y}]=-2y'\) (not merely the reduced Case‑2 separable equation): implicitly differentiating the relation in Sympy and substituting the resulting \(y'\) back into \(x[(y')^2+e^{2y}]+2y'\) gives \(0\) exactly (checked both symbolically and by numerical spot-check). So this family is confirmed correct.

5. The companion family from the other branch

The source only ever solved the "\(-\)" branch (Case 2). Repeating the identical substitution on the "\(+\)" branch (\(u' = u(-1+\sqrt{1-x^2u^2})/x\)) gives \(dv/dx = +\frac{v}{x}\sqrt{1-v^2}\), i.e. the same separable equation with the sign of \(dx/x\) flipped. Integrating gives \(\sqrt{1-v^2}+1 = K v/x = Ku\), i.e. \[ \sqrt{1-x^2e^{2y}}+1 = Ke^{y}. \] This companion family was likewise verified with Sympy (implicit differentiation, substitution into the original quadratic gives \(0\) at a numerical test point to machine precision). So the full solution of the original equation consists of both families, one per sign of the quadratic in \(y'\); the source only presented one of the two.