V 13.2 DSolve now returns {} as solution for some ode's, while V 13.1 returns solutions for same ode's but with Solve in the solution (i.e. implicit solutions).
The question is: why this change and is there a way/option to make DSolve return those implicit solutions instead of a no solution? I think implicit solution is better than no solution.
Or is it possible that those solutions returned by V 13.1 were wrong and that is why V 13.2 no longer returns them?
The following are two examples, I can find more if possible.
ode = (y'[x])^2 == Exp[4*x - 2*y[x]]*(y'[x] - 1)
DSolve[ode, y[x], x]
ode = 4*(y'[x])^2 + 2*Exp[2*x - 2*y[x]]*y'[x] - Exp[2*x - 2*y[x]] == 0
DSolve[ode, y[x], x]
The following is the output from V 13.1
Update
fyi, here is my hand solution to the first ode, verified by Mathematica
\begin{align*} \left( y^{\prime}\right) ^{2} & =e^{4x-2y}\left( y^{\prime}-1\right) \\ \ln\left( y^{\prime}\right) ^{2} & =\left( 4x-2y\right) +\ln\left( y^{\prime}-1\right) \\ 4x-2y & =\ln\left( y^{\prime}\right) ^{2}-\ln\left( y^{\prime}-1\right) \\ 4x-2y & =\ln\frac{\left( y^{\prime}\right) ^{2}}{y^{\prime}-1}\\ 2y & =4x-\ln\frac{\left( y^{\prime}\right) ^{2}}{y^{\prime}-1}\\ y & =2x-\frac{1}{2}\ln\left( \frac{\left( y^{\prime}\right) ^{2}% }{y^{\prime}-1}\right) \\ & =\phi\left( x,y^{\prime}\right) \end{align*} Let $y^{\prime}=p$ \begin{align} y & =2x-\frac{1}{2}\ln\left( \frac{p^{2}}{p-1}\right) \tag{1}\\ & =\phi\left( x,p\right) \nonumber \end{align} The above is d'Alembert ode. Differentiating w.r.t. $x$ gives \begin{align} y^{\prime} & =\frac{\partial\phi}{\partial x}+\frac{\partial\phi}{\partial p}\frac{dp}{dx}\nonumber\\ p & =2+\left( \frac{\left( \frac{2p}{p-1}-\frac{p^{2}}{\left( p-1\right) ^{2}}\right) \left( 1-p\right) }{2p^{2}}\right) \frac{dp}{dx}\nonumber\\ p-2 & =\left( \frac{2-p}{2p^{2}-2p}\right) \frac{dp}{dx}\tag{2} \end{align} The singular solution is when $\frac{dp}{dx}=0$ which gives $p=2$. From (1) this gives $$ y=2x-\frac{1}{2}\ln4 $$ The general solution is when $\frac{dp}{dx}\neq0$. Then (2) becomes \begin{align*} \frac{dp}{dx} & =\left( p-2\right) \left( \frac{2p^{2}-2p}{2-p}\right) \\ & =2p\left( 1-p\right) \end{align*} The above is separable. Solving for $p$ gives $$ p=\frac{1}{1+ce^{-2x}} $$ Substituting the above solutions of $p$ in (1) gives \begin{align*} y & =2x-\frac{1}{2}\ln\left( \frac{\left( \frac{1}{1+ce^{-2x}}\right) ^{2}}{\frac{1}{1+ce^{-2x}}-1}\right) \\ & =2x-\frac{1}{2}\ln\left( \frac{-e^{4x}}{c\left( c+e^{2x}\right) }\right) \end{align*}
Verification
ClearAll[y,x];
ode = (y'[x])^2 == Exp[4*x - 2*y[x]]*(y'[x] - 1);
mysol=y->Function[{x},2*x-1/2*Log[-Exp[4*x]/(C[1]*(C[1]+Exp[2*x]))]];
ode/.mysol//Simplify
(*True*)
mysol2=y->Function[{x},2*x-1/2*Log[4]]
ode/.mysol2//Simplify
(*True*)
