Solving second-order nonlinear autonomous differential equations For some physical purposes, I am now interested in differential equations of the form $$y''=f(y)$$
Now, this resource gives a general solution to this equation (sans constant solutions) as $$\int dy\left(C_1+2\int dy \,f(y)\right)^{-1/2}=C_2\pm x$$
but I have no idea how this formula was derived, and I can't find any resources on it other than the above; is this formula correct, and if so, how is it derived?
 A: Hint:
$$y''=f(y)$$
Multiply by $2y′$ both sides:
$$2y'y''=2f(y)y'$$
Integrate to reduce the order.
$$y'^2=2\int f(y) dy +C_1$$
It's separable.
A: Starting from
$y'' = f(y), \tag 1$
we multiply through by $y'$:
$y''y' = f(y)y'; \tag 2$
we observe that
$\dfrac{1}{2}((y')^2)' = \dfrac{1}{2} (2y'y'') = y'y'', \tag 3$
whence (2) may be written
$\dfrac{1}{2}((y')^2)' = f(y)y', \tag 4$
or
$((y')^2)' = 2f(y)y'; \tag 5$
we integrate with respect to $x$:
$(y')^2 = \displaystyle \int ((y')^2)' \; dx + C_1$
$= 2 \displaystyle \int f(y)y' \; dx + C_1 = 2 \int f(y) \; dy + C_1; \tag 6$
we isolate $y'$:
$\dfrac{dy}{dx} = y' = \pm \left (2 \displaystyle \int f(y) \; dy + C_1 \right)^{1/2}, \tag 7$
whence
$\dfrac{dx}{dy} = \pm \left (2 \displaystyle \int f(y) \; dy + C_1 \right )^{-1/2}; \tag 8$
we may now integrate again over $y$:
$x + C_2 = \displaystyle \int \dfrac{dx}{dy} \; dy = \pm \int \left (2 \int f(y) \; dy + C_1\right )^{-1/2} \; dy, \tag 9$
or
$\pm(x + C_2) = \displaystyle \int \dfrac{dx}{dy} \; dy = \int \left (2 \int f(y) \; dy + C_1\right )^{-1/2} \; dy; \tag{10}$
we may absorb the $\pm$ sign into $C_2$ since it is an arbitrary constant, and write
$\pm x + C_2 = \displaystyle \int \dfrac{dx}{dy} \; dy = \int \left (2 \int f(y) \; dy + C_1\right )^{-1/2} \; dy, \tag{11}$
which is essentially identical to the formula given by our OP John Dumancic in the body of the question itself.
A: There is another way to do it.
Switch variables and write the equation as
$$-\frac{x''}{(x')^3}=f(y)$$ Now, reduction of order $p=x'$ to give
$$-\frac{p'}{p^3}=f(y)\implies \frac{1}{2p^2}=\int f(y)\,dy=F(y)+C\implies p=\frac{\pm1}{\sqrt{C_1+2 F(y)}}$$
$$x=\pm \int \frac{dy}{\sqrt{C_1+2 F(y)}}+C_2$$
