# Cases of Reduction of Order

The differential equation of the $$n$$th order in the general case has the form:

$F\left( {x,y,y',y^{\prime\prime}, \ldots ,{y^{\left( n \right)}}} \right) = 0,$

where $$F$$ is a continuous function of the specified arguments.

The order of the equation can be reduced if it does not contain some of the arguments, or has a certain symmetry. Below we consider in detail some cases of reducing the order with respect to the differential equations of arbitrary order $$n.$$ Transformation of the $$2$$nd order equations is described here.

## Case $$1.$$ Equation of Type $$F\left( {x,{y^{\left( k \right)}},{y^{\left( {k + 1} \right)}}, \ldots ,{y^{\left( n \right)}}} \right)$$ $$= 0$$

If the differential equation does not contain the original function and its $$k - 1$$ first derivatives, then by replacing

${y^{\left( k \right)}} = p\left( x \right)$

the order of this equation is reduced by $$k$$ units. As a result, the original equation takes the form

$F\left( {x,p,p', \ldots {p^{\left( {n - k} \right)}}} \right) = 0.$

From this equation (if possible) we can determine the function $$p\left( x \right).$$ The original function $$y\left( x \right)$$ can be found by $$k$$-fold integration.

If the differential equation does not contain only the original function $$y,$$ that is has the form

$F\left( {x,y',y^{\prime\prime}, \ldots ,{y^{\left( n \right)}}} \right) = 0,$

then its order can be reduced by one by the substitution $$y = p\left( x \right).$$

## Case $$2.$$ Equation of Type $$F\left( {y,y',y^{\prime\prime}, \ldots ,{y^{\left( n \right)}}} \right) = 0$$

Here the left side does not contain the independent variable $$x.$$ The order of the equation can be reduced by the substitution $$y = p\left( y \right).$$ The derivatives are defined through the new variables $$y$$ and $$p$$ as follows:

$y' = \frac{{dy}}{{dx}} = p,$
$y^{\prime\prime} = \frac{{{d^2}y}}{{d{x^2}}} = \frac{d}{{dx}}\left( {\frac{{dy}}{{dx}}} \right) = \frac{{dp}}{{dx}} = \frac{{dp}}{{dy}}\frac{{dy}}{{dx}} = p\frac{{dp}}{{dy}},$
$y^{\prime\prime\prime} = \frac{{{d^3}y}}{{d{x^3}}} = \frac{d}{{dx}}\left( {p\frac{{dp}}{{dy}}} \right) = \frac{d}{{dy}}\left( {p\frac{{dp}}{{dy}}} \right)\frac{{dy}}{{dx}} = \left[ {p\frac{{{d^2}p}}{{d{y^2}}} + {{\left( {\frac{{dp}}{{dy}}} \right)}^2}} \right]p = {p^2}\frac{{{d^2}p}}{{d{y^2}}} + p{\left( {\frac{{dp}}{{dy}}} \right)^2},$
${y^{IV}} = \frac{{{d^4}y}}{{d{x^4}}} = \frac{d}{{dx}}\left[ {{p^2}\frac{{{d^2}p}}{{d{y^2}}} + p{{\left( {\frac{{dp}}{{dy}}} \right)}^2}} \right] = \frac{d}{{dy}}\left[ {{p^2}\frac{{{d^2}p}}{{d{y^2}}} + p{{\left( {\frac{{dp}}{{dy}}} \right)}^2}} \right]\frac{{dy}}{{dx}} = \left[ {{p^2}\frac{{{d^3}p}}{{d{y^3}}} + 2p\frac{{dp}}{{dy}}\frac{{{d^2}p}}{{d{y^2}}} + 2p\frac{{dp}}{{dy}}\frac{{{d^2}p}}{{d{y^2}}} + {{\left( {\frac{{dp}}{{dy}}} \right)}^3}} \right]p = {p^3}\frac{{{d^3}p}}{{d{y^3}}} + 4{p^2}\frac{{dp}}{{dy}}\frac{{{d^2}p}}{{d{y^2}}} + p{\left( {\frac{{dp}}{{dy}}} \right)^3}.$

It is seen that substitution of the derivatives into the original equation gives a new differential equation of the $$\left( {n - 1} \right)$$th order. Solving this equation, we can determine the function $$p\left( y \right)$$ and then find $$y\left( x \right).$$

## Case $$3.$$ Homogeneous Equation $$F\left( {x,y,y',y^{\prime\prime}, \ldots ,{y^{\left( n \right)}}} \right) = 0$$

The equation $$F\left( {x,y,y',y^{\prime\prime}, \ldots ,}\right.$$ $$\left.{{y^{\left( n \right)}}} \right) = 0$$ is called homogeneous with respect to the arguments $${y,y',}$$ $${y^{\prime\prime}, \ldots ,}$$ $${{y^{\left( n \right)}}}$$ if the following identity holds:

$F\left( {x,ky,ky',ky^{\prime\prime}, \ldots ,k{y^{\left( n \right)}}} \right) \equiv {k^m}F\left( {x,y,y',y^{\prime\prime}, \ldots ,{y^{\left( n \right)}}} \right).$

The order of this equation can be reduced by one using the substitution

$y = {e^{\int {zdx} }},$

where $$z\left( x \right)$$ is the new unknown function.

After $$z\left( x \right)$$ is determined, we can find the original function $$y\left( x \right)$$ by integration using the formula

$y\left( x \right) = {C_1}{e^{\int {zdx} }},$

where $${C_1}$$ is an arbitrary number.

## Case $$4.$$ Function $$F\left( {x,y,y',y^{\prime\prime}, \ldots ,{y^{\left( n \right)}}} \right)$$ is a Total Derivative

In some cases, the left-hand side $$F\left( {x,y,y',y^{\prime\prime}, \ldots ,{y^{\left( n \right)}}} \right)$$ of the differential equation can be expressed as the total derivative with respect to $$x$$ of a differential expression of the $$\left( {n - 1} \right)$$th order:

$F\left( {x,y,y',y^{\prime\prime}, \ldots ,{y^{\left( n \right)}}} \right) = \frac{d}{{dx}}\Phi \left( {x,y,y',y^{\prime\prime}, \ldots ,{y^{\left( {n - 1} \right)}}} \right).$

Then the solution of the original equation can be written as

$\Phi \left( {x,y,y',y^{\prime\prime}, \ldots ,{y^{\left( {n - 1} \right)}}} \right) = C,$

where $$C$$ is an arbitrary constant.