# 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?

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.

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.

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 $$-\frac{p'}{p^3}=f(y)\implies \frac{1}{2p^2}=\int f(y)\,dy=F(y)+C\implies p=\frac{\pm1}{\sqrt{C_1+2 F(y)}}$$ $$x=\pm \int \frac{dy}{\sqrt{C_1+2 F(y)}}+C_2$$