The second order ordinary differential equation

$\displaystyle\frac{d^{2}y}{dx^{2}}\;=\;f\!\left(x,\,y,\,\frac{dy}{dx}\right)$ | (1) |

may in certain special cases be solved by using two quadratures, sometimes also by reduction to a first order differential equation and a quadrature.

If the right hand side of (1) contains at most one of the quantities $x$, $y$ and $\frac{dy}{dx}$, the general solution solution is obtained by two quadratures.

- •
- •
The equation

$\displaystyle\frac{d^{2}y}{dx^{2}}\,=\,f(y)$ (3) has as constant solutions all real roots of the equation $f(y)=0$. The other solutions can be gotten from the normal system

$\displaystyle\frac{dy}{dx}\,=\,z,\qquad\frac{dz}{dx}\,=\,f(y)$ (4) of (3). Dividing the equations (4) we get now $\frac{dz}{dy}=\frac{f(y)}{z}$. By separation of variables and integration we may write

$\frac{z^{2}}{2}=\int\!f(y)\,dy+C_{1},$ whence the first equation of (4) reads

$\frac{dy}{dx}\,=\,\sqrt{2\!\int\!f(y)\,dy+C_{1}}.$ Separating here the variables and integrating give the general integral of (3) in the form

$\displaystyle\int\!\frac{dy}{\sqrt{2\!\int\!f(y)\,dy+C_{1}}}\;=\;x+C_{2}.$ (5) The integration constant $C_{1}$ has an influence on the form of the integral curves, but $C_{2}$ only translates them in the direction of the $x$-axis.

- •
The equation

$\displaystyle\frac{d^{2}y}{dx^{2}}\,=\,f(\frac{dy}{dx})$ (6) is equivalent with the normal system

$\displaystyle\frac{dy}{dx}\,=\,z,\quad\frac{dz}{dx}\,=\,f(z).$ (7) If the equation $f(z)=0$ has real roots $z_{1},\,z_{2},\,\ldots$, these satisfy the latter of the equations (7), and thus, according to the former of them, the differential equation (6) has the solutions $y:=z_{1}x+C_{1}$, $y:=z_{2}x+C_{2},\;\ldots$.

The other solutions of (6) are obtained by separating the variables and integrating:

$\displaystyle x\,=\,\int\!\frac{dz}{f(z)}+C.$ (8) If this antiderivative is expressible in closed form and if then the equation (8) can be solved for $z$, we may write

$z\,=\,\frac{dy}{dx}\,=\,g(x\!-\!C).$ Accordingly we have in this case the general solution of the ODE (6):

$\displaystyle y\;=\;\int\!g(x\!-\!C)\,dx+C^{\prime}.$ (9) In other cases, we express also $y$ as a function of $z$. By the chain rule, the normal system (7) yields

$\frac{dy}{dz}\,=\,\frac{dy}{dx}\cdot\frac{dx}{dz}\,=\,\frac{z}{f(z)},$ whence

$y=\int\frac{z\,dz}{f(z)}+C^{\prime}.$ Thus the general solution of (6) reads now in a parametric form as

$\displaystyle x\,=\,\int\!\frac{dz}{f(z)}+C,\qquad y=\int\frac{z\,dz}{f(z)}+C^% {\prime}.$ (10) The equations 10 show that a translation of any integral curve yields another integral curve.