Find the general solution of (y')^2 = y''
Find the general solution of $(y')^2 = y''$.
Explanation of Method
Reduce the order via $p = y'$, then integrate twice.
Step 1: Substitution
Let $p = y'$, so $p' = y''$. The equation becomes: $p^2 = p'.$
This is a first-order separable ODE in $p(x)$.
Step 2: Solve for $p$
Case 1 · p = 0.
Then $y' = 0$, giving the trivial solution $y = C$, a constant.
Case 2 · p ≠ 0.
Separate variables: $\frac{dp}{p^2} = dx.$
Integrate both sides: \[\int p^{-2}\,dp = \int dx,\qquad -\frac{1}{p} = x + C_1,\qquad p = -\frac{1}{x + C_1}.\]
Thus: \[y' = -\frac{1}{x + C_1}. \tag{1}\]
Step 3: Integrate to find $y$
Integrate (1): $y = \int -\frac{1}{x+C_1}\,dx = -\ln|x + C_1| + C_2.$
Step 4: General solution
The solution is recorded in the result below.
Step 5: Verification
Differentiate: $y' = -\frac{1}{x+C_1},\qquad y'' = \frac{1}{(x+C_1)^2}.$
Check: $(y')^2 = \frac{1}{(x+C_1)^2} = y''.$ $\checkmark$
Result
Final answer:
\[\boxed{\,y = C_2 - \ln|x + C_1|\,}\]
Remark
Domain. $C_1$ and $C_2$ are arbitrary constants, valid for $x \neq -C_1$.