# Cases of Reduction of Order

The differential equation of the nth 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 2nd 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.

See solved problems on Page 2.