Solving second-order nonlinear autonomous differential equations

Question

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?

Answer 1

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.

Answer 2

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.

Answer 3

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