Find the general and singular solutions (if any) of…

Find the general and singular solutions (if any) of $(y'+1)^3=27(x+y)^2$.

Step 1: Substitute $u=x+y$

$u'=1+y'$, so $y'+1=u'$ and the equation becomes $(u')^3=27u^2$, i.e. $u'=3u^{2/3}$.

Step 2: Solve for $u\ne0$

$u^{-2/3}\,du=3\,dx\Rightarrow3u^{1/3}=3x+C_1\Rightarrow u=(x+C)^3$ (with $C=C_1/3$).

Step 3: Return to $y$

Since $u=x+y$: $y=(x+C)^3-x$.

Step 4: $p$-discriminant and the degenerate case $u=0$

With $F(x,y,p)=(p+1)^3-27(x+y)^2$: $F_p=3(p+1)^2=0\Rightarrow p=-1$; substituting gives $27(x+y)^2=0$, i.e. $y=-x$. Directly: $y=-x\Rightarrow y'=-1$, and $(y'+1)^3=0=27(x+y)^2=27\cdot0^2$: satisfied.

Step 5: Confirm $y=-x$ is a genuine envelope

Eliminating $C$ from $G=y-(x+C)^3+x=0,\ G_C=-3(x+C)^2=0$ gives $C=-x$, and substituting back gives $y=-x$ — matching the $p$-discriminant result. Since $y=-x$ cannot equal $(x+C)^3-x$ for any single fixed $C$ over an interval, it is a genuine singular solution.