ODE types solvable by two quadratures
The second order ordinary differential equation
$\frac{{d}^{2}y}{d{x}^{2}}}=f(x,y,{\displaystyle \frac{dy}{dx}})$  (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.

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The equation
$\frac{{d}^{2}y}{d{x}^{2}}}=f(x)$ (2) is considered here (http://planetmath.org/EquationYFx).

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The equation
$\frac{{d}^{2}y}{d{x}^{2}}}=f(y)$ (3) has as constant solutions all real roots of the equation $f(y)=0$. The other solutions can be gotten from the normal system
$\frac{dy}{dx}}=z,{\displaystyle \frac{dz}{dx}}=f(y)$ (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
$$\frac{{z}^{2}}{2}=\int f(y)\mathit{d}y+{C}_{1},$$ whence the first equation of (4) reads
$$\frac{dy}{dx}=\sqrt{2\int f(y)\mathit{d}y+{C}_{1}}.$$ here the variables and integrating give the general integral of (3) in the form
$\int \frac{dy}{\sqrt{2\int f(y)\mathit{d}y+{C}_{1}}}}=x+{C}_{2}.$ (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.

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The equation
$\frac{{d}^{2}y}{d{x}^{2}}}=f({\displaystyle \frac{dy}{dx}})$ (6) is equivalent^{} (http://planetmath.org/Equivalent3) with the normal system
$\frac{dy}{dx}}=z,{\displaystyle \frac{dz}{dx}}=f(z).$ (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
$$z=\frac{dy}{dx}=g(xC).$$ Accordingly we have in this case the general solution of the ODE (6):
$y={\displaystyle \int g(xC)\mathit{d}x}+{C}^{\prime}.$ (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.
Title  ODE types solvable by two quadratures 

Canonical name  ODETypesSolvableByTwoQuadratures 
Date of creation  20150320 17:04:58 
Last modified on  20150320 17:04:58 
Owner  pahio (2872) 
Last modified by  pahio (2872) 
Numerical id  13 
Author  pahio (2872) 
Entry type  Topic 
Classification  msc 34A05 
Synonym  second order ODE types solvable by quadratures 
Related topic  ODETypesReductibleToTheVariablesSeparableCase 