# ODE types solvable by two quadratures

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

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

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

• 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

 $\frac{dy}{dx}\,=\,\sqrt{2\!\int\!f(y)\,dy+C_{1}}.$

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

 $\displaystyle\frac{d^{2}y}{dx^{2}}\,=\,f(\frac{dy}{dx})$ (6)
 $\displaystyle\frac{dy}{dx}\,=\,z,\quad\frac{dz}{dx}\,=\,f(z).$ (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.

Title ODE types solvable by two quadratures ODETypesSolvableByTwoQuadratures 2015-03-20 17:04:58 2015-03-20 17:04:58 pahio (2872) pahio (2872) 13 pahio (2872) Topic msc 34A05 second order ODE types solvable by quadratures ODETypesReductibleToTheVariablesSeparableCase