Solve displaystyle y'=(1)/(2y)·(2x+3y^2+1)/(3x+4y^2+1)

Solve $\displaystyle y'=\frac{1}{2y}\cdot\frac{2x+3y^2+1}{3x+4y^2+1}$.

Step 1: Confirm the setup through the homogeneous equation

\texttt{sympy.solve} on $\{2h+3k+1=0,\ 3h+4k+1=0\}$ returns $h=1,\ k=-1$, matching the source. With $v=y^2$, $x=X+1$, $v=V-1$, the equation becomes $\dfrac{dV}{dX}=\dfrac{2X+3V}{3X+4V}$, and with $V=uX$, \texttt{sympy} confirms $X\dfrac{du}{dX}=\dfrac{2-4u^2}{3+4u}$, i.e. \[\frac{3+4u}{2-4u^2}\,du=\frac{dX}{X},\] exactly as the source has it — the setup is entirely sound.

Step 2: Redo the integral cleanly with \texttt{sympy} (the source's flawed step)

Calling \texttt{sp.integrate}$\!\left(\dfrac{3+4u}{2-4u^2},u\right)$ directly returns $-\dfrac{(4+3\sqrt2)}{8}\ln\!\left(u-\tfrac{\sqrt2}{2}\right)-\dfrac{(4-3\sqrt2)}{8}\ln\!\left(u+\tfrac{\sqrt2}{2}\right)$, which regroups into clean log terms as \[\int\frac{3+4u}{2-4u^2}\,du=\frac{3}{4\sqrt2}\ln\left|\frac{1+\sqrt2\,u}{1-\sqrt2\,u}\right|-\frac12\ln\left|1-2u^2\right|+\text{const},\] confirmed by \texttt{sympy}: differentiating this expression reproduces $\dfrac{3+4u}{2-4u^2}$ exactly (residual $0$). Comparing to the source: its arctanh/log term $\dfrac{3}{4\sqrt2}\ln\left|\dfrac{1+\sqrt2u}{1-\sqrt2u}\right|$ was actually correct, but its second term, $-\ln|1-2u^2|$, is wrong — it is missing a factor of $\tfrac12$ (the source integrated $\int\frac{2u}{1-2u^2}\,du$ as $-\ln|1-2u^2|$ instead of the correct $-\tfrac12\ln|1-2u^2|$), a genuine arithmetic slip, not merely an abandoned computation.

Step 3: Back-substitute and state the corrected implicit solution

With $u=V/X=(v-1)/(x-1)=(y^2+1)/(x-1)$ (translating back through $v=y^2$), and absorbing $\ln|X|=\ln|x-1|$: \[\frac{3}{4\sqrt2}\ln\left|\frac{1+\sqrt2\,u}{1-\sqrt2\,u}\right|-\frac12\ln\left|1-2u^2\right|=\ln|x-1|+C,\qquad u=\frac{y^2+1}{x-1}.\]

Step 4: Verify by implicit differentiation from scratch

Substituting $u=(y^2+1)/(x-1)$ into the corrected implicit relation and implicit-differentiating with \texttt{sympy} (solving $\partial F/\partial x+\partial F/\partial y\cdot y'=0$ for $y'$) gives, after simplification, \[y'=\frac{2x+3y^2+1}{2y(3x+4y^2+1)},\] which is \texttt{sympy}-confirmed identical to the original equation $y'=\dfrac{1}{2y}\cdot\dfrac{2x+3y^2+1}{3x+4y^2+1}$ — residual exactly $0$. (Re-differentiating the source's original, uncorrected boxed expression instead gives a residual that does not vanish, confirming the source's final answer was genuinely wrong, not just inelegantly presented.)