# Differential Equations

## Higher Order Equations # 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:

$y’ = \frac{{dy}}{{dx}} = p,$

${y^{\prime\prime} = \frac{{{d^2}y}}{{d{x^2}}} = \frac{d}{{dx}}\left( {\frac{{dy}}{{dx}}} \right) } = {\frac{{dp}}{{dx}} } = {\frac{{dp}}{{dy}}\frac{{dy}}{{dx}} } = {p\frac{{dp}}{{dy}},}$

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

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

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.

Consider examples for the various cases of order reduction.

## Solved Problems

Click or tap a problem to see the solution.

### Example 1

Find the general solution of the differential equation $$y^{\prime\prime\prime} + {\large\frac{2}{x}\normalsize} y^{\prime\prime} = 0.$$

### Example 2

Find the general solution of the equation $$5{\left( {y^{\prime\prime\prime}} \right)^2} – 3y^{\prime\prime}{y^{IV}} = 0.$$

### Example 3

Find the general solution of the equation $$y’y^{\prime\prime\prime} + {\left( {y^{\prime\prime}} \right)^2} = 0.$$

### Example 4

Find the general solution of the differential equation $$yy^{\prime\prime\prime} – y’y^{\prime\prime} = 0.$$
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Concept
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Problems 1-4