ODE types solvable by two quadratures

The second order ordinary differential equation

${\displaystyle \frac{d^{2}y}{d{x}^2}}=f(x,y,{\displaystyle \frac{dy}{dx}})$(1)

may in certain special cases be solved by using two quadratures, sometimes also by reduction to a first order differential equation (http://planetmath.org/ODE) 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

  ${\displaystyle \frac{d^{2}y}{d{x}^2}}=f(x)$(2)

  is considered here (http://planetmath.org/EquationYFx).
  

• The equation

  ${\displaystyle \frac{d^{2}y}{d{x}^2}}=f(y)$(3)

  has as constant solutions all real roots of the equation  $f(y)=0$.  The other solutions can be gotten from the normal system

  ${\displaystyle \frac{dy}{dx}}=z,{\displaystyle \frac{dz}{dx}}=f(y)$(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

  $$\frac{dy}{dx}=\sqrt{2\int f(y)𝑑y+C_1}.$$

  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 (http://planetmath.org/SolutionsOfOrdinaryDifferentialEquation) $C_1$ 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

  ${\displaystyle \frac{d^{2}y}{d{x}^2}}=f({\displaystyle \frac{dy}{dx}})$(6)

  is equivalent (http://planetmath.org/Equivalent3) with the normal system

  ${\displaystyle \frac{dy}{dx}}=z,{\displaystyle \frac{dz}{dx}}=f(z).$(7)

  If the equation  $f(z)=0$  has real roots  $z_1,z_2,\mathrm{\dots }$,  these satisfy the latter of the equations (7), and thus, according to the former of them, the differential equation (6) has the solutions  $y:=z_{1}x+C_1$,  $y:=z_{2}x+C_2,\mathrm{\dots }$.

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

  $x={\displaystyle \int \frac{dz}{f(z)}}+C.$(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):

  $y={\displaystyle \int g(x-C)𝑑x}+C^{\prime }.$(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

  $$y=\int \frac{zdz}{f(z)}+C^{\prime }.$$

  Thus the general solution of (6) reads now in a parametric form as

  $x={\displaystyle \int \frac{dz}{f(z)}}+C,y={\displaystyle \int \frac{zdz}{f(z)}}+C^{\prime }.$(10)

  The equations 10 show that a translation of any integral curve yields another integral curve.