Reduction of Order

A second-order equation is generally much harder than a first-order one. But if the equation is missing either the dependent variable $y$ or the independent variable $x$ , a substitution lowers its order by one, and the tools of this chapter finish the job. This is the last technique of the chapter and the natural bridge to the higher order chapters that follow.

Quick Reference

Case	Equation shape	Substitution	Result
$y$ missing	F\left(x, y', y''\right) = 0	$p = \frac{d y}{d x}$ , so y'' = \dfrac{dp}{dx}	First-order equation in $p$ and $x$
$x$ missing	F\left(y, y', y''\right) = 0	$p = \frac{d y}{d x}$ , so y'' = p\dfrac{dp}{dy}	First-order equation in $p$ and $y$
After solving for $p$	Replace $p$ by $\frac{d y}{d x}$	Integrate once more	Two arbitrary constants total
Chain rule identity	$\frac{d^{2} y}{d x^{2}} = \frac{d p}{d x} = \frac{d p}{d y} \frac{d y}{d x} = p \frac{d p}{d y}$		The key to Case 2

The Setting

The most general form of a second-order differential equation is

F\left(x, y, y', y''\right) = 0 .

Two special types of second-order equations can be reduced to first-order equations: those in which $y$ does not appear, and those in which $x$ does not appear. In both cases the trick is to make y' the new unknown.

Case 1: The Dependent Variable Does Not Appear

Suppose the equation has the form

F\left(x, y', y''\right) = 0 \tag{1}

so that $y$ itself is absent, though $x$ , y', and y'' may all appear.

To solve this equation, introduce a new dependent variable $p$ :

p = \frac{d y}{d x} and \frac{d p}{d x} = \frac{d^{2} y}{d x^{2}} .

If we replace y' and y'' in (1) by $p$ and p', we get

G\left(x, p, \frac{dp}{dx}\right) = 0 . \tag{2}

This is a first-order equation in the unknown $p$ , and the whole of this chapter applies to it. If we can solve (2), we then replace $p$ in the solution by y' and integrate once more to recover $y$ .

The two integrations produce two arbitrary constants, which is the right number for a second-order equation. The first constant enters when solving the first-order equation for $p$ ; the second enters when integrating $p$ to get $y$ .

Solve

x y'' + 2y' = x + 1 .

Solution

The variable $y$ does not appear, so this is Case 1. Let p = y' and p' = y''. Then

x y'' + 2y' = x + 1 \quad \Longrightarrow \quad x p' + 2p = x + 1

or, dividing by $x$ ,

\frac{d p}{d x} + \frac{2}{x} p = 1 + \frac{1}{x}

which is a linear first-order differential equation. The integrating factor is

e^{\int \frac{2}{x} d x} = e^{2 \ln x} = e^{\ln x^{2}} = x^{2} .

Therefore

\begin{aligned} x^{2}\left(\frac{dp}{dx} + \frac{2}{x}p\right) &= x^{2} + x \\ \frac{d}{dx}\left(x^{2}p\right) &= x^{2} + x \\ x^{2}p &= \frac{1}{3}x^{3} + \frac{1}{2}x^{2} + c \end{aligned}

Dividing by $x^{2}$ ,

p = \frac{1}{3} x + \frac{1}{2} + \frac{c}{x^{2}} .

But $p = \frac{d y}{d x}$ , so

\frac{d y}{d x} = \frac{1}{3} x + \frac{1}{2} + c x^{- 2} .

Now $y$ is obtained by one more integration, using $\int x^{- 2} d x = - x^{- 1}$ :

y = \frac{1}{6} x^{2} + \frac{1}{2} x - c x^{- 1} + k .

Letting $c_{1} = - c$ and $c_{2} = k$ , the solution is

y = \frac{1}{6} x^{2} + \frac{1}{2} x + \frac{c_{1}}{x} + c_{2} .

Two arbitrary constants, as expected for a second-order equation.

We want to find the position of a falling ball as a function of time $t$ , taking air resistance into account. Taking the air resistance force to be proportional to the square of the velocity, the equation of motion becomes

m \frac{d^{2} s}{d t^{2}} = m g - k {(\frac{d s}{d t})}^{2}

where $s (t)$ is the distance fallen at time $t$ , $m$ is the mass of the body, $g$ is the acceleration due to gravity, and $k$ is a constant. Suppose the ball starts from rest at $t = 0$ .

Solution

The position $s$ does not appear in the equation, only its derivatives, so this is Case 1. Let

v = \frac{d s}{d t} and \frac{d v}{d t} = \frac{d^{2} s}{d t^{2}} .

Then the governing equation becomes

m \frac{d v}{d t} = m g - k v^{2},

a first-order equation for the velocity. It is separable:

\frac{m d v}{m g - k v^{2}} = d t .

Factor $m g$ out of the denominator:

\frac{m}{m g} \int \frac{d v}{1 - \frac{k}{m g} v^{2}} = t + c .

To integrate, make the substitution

u = \sqrt{\frac{k}{m g}} v ⟹ d u = \sqrt{\frac{k}{m g}} d v ⟹ d v = \sqrt{\frac{m g}{k}} d u .

Therefore

\frac{1}{g} \int \frac{1}{1 - u^{2}} \underset{d v}{\underset{⏟}{\sqrt{\frac{m g}{k}} d u}} = \sqrt{\frac{m}{g k}} \int \frac{d u}{1 - u^{2}} .

Because partial fractions give

\frac{1}{1 - u^{2}} = \frac{\frac{1}{2}}{1 - u} + \frac{\frac{1}{2}}{1 + u},

we have

\begin{aligned} \frac{1}{2}\sqrt{\frac{m}{gk}}\int\left(\frac{1}{1-u} + \frac{1}{1+u}\right)du &= \frac{1}{2}\sqrt{\frac{m}{gk}}\left[-\ln(1-u) + \ln(1+u)\right] \\ &= \frac{1}{2}\sqrt{\frac{m}{gk}}\,\ln\left(\frac{1+u}{1-u}\right) \end{aligned}

Putting back $u = \sqrt{\frac{k}{m g}} v$ ,

\frac{1}{2} \sqrt{\frac{m}{g k}} \ln (\frac{1 + \sqrt{\frac{k}{m g}} v}{1 - \sqrt{\frac{k}{m g}} v}) = t + c .

Because $v = 0$ at $t = 0$ , the logarithm's argument is $1$ and $\ln 1 = 0$ , so $c = 0$ . Thus

t = \frac{1}{2} \sqrt{\frac{m}{g k}} \ln \frac{1 + \sqrt{\frac{k}{m g}} v}{1 - \sqrt{\frac{k}{m g}} v} .

Since

\tanh^{- 1} x = \frac{1}{2} \ln \frac{1 + x}{1 - x},

the result can be written as

t = \sqrt{\frac{m}{g k}} \tanh^{- 1} (\sqrt{\frac{k}{m g}} v)

or, solving for $v$ ,

v = \sqrt{\frac{m g}{k}} \tanh (\sqrt{\frac{g k}{m}} t) .

Since $- 1 < \tanh x < 1$ , the speed of the falling ball never exceeds $\sqrt{\frac{m g}{k}}$ . This limiting speed is the terminal velocity $v_{T}$ :

v_{T} = \sqrt{\frac{m g}{k}} .

Note that $\sqrt{\frac{g k}{m}} = \frac{g}{v_{T}}$ , so the velocity can be written compactly as

v = v_{T} \tanh (\frac{g t}{v_{T}}) .

Now integrate the velocity to find the position:

s (t) = \int_{0}^{t} v (τ) d τ .

Since $\tanh x = \frac{\sinh x}{\cosh x}$ and $\frac{d}{d x} \cosh x = \sinh x$ , we have $\int \tanh (a x) d x = \frac{1}{a} \ln (\cosh (a x)) + c$ . Here $a = g / v_{T}$ , so

s (t) = v_{T} \cdot \frac{v_{T}}{g} \ln (\cosh \frac{g t}{v_{T}}) + c .

Assuming $s (0) = 0$ , and noting that $\cosh 0 = 1$ so the logarithm vanishes, $c = 0$ and the position function is

s (t) = \frac{v_{T}^{2}}{g} \ln (\cosh \frac{g t}{v_{T}}) .

Sanity check for large $t$ . For large arguments $\cosh z \approx \frac{1}{2} e^{z}$ , so $s (t) \approx \frac{v_{T}^{2}}{g} (\frac{g t}{v_{T}} - \ln 2) = v_{T} t - \frac{v_{T}^{2} \ln 2}{g}$ . The ball eventually falls at the constant terminal velocity, lagging behind a drag-free fall by a fixed distance, exactly as physical intuition demands.

Case 2: The Independent Variable Does Not Appear

Now suppose the equation has the form

F\left(y, y', y''\right) = 0

so that $x$ is absent. Such an equation is called autonomous.

Setting p = y' is again the right move, but now the natural variable to keep is $y$ , not $x$ . The chain rule supplies the necessary identity:

Substituting y' = p and y'' = p\dfrac{dp}{dy} turns the equation into a first-order equation for $p$ as a function of $y$ . Solve it, then recover $y$ from the separable equation $\frac{d y}{d x} = p (y)$ .

Solve y y'' + \left(y'\right)^{2} = 0.

Solution

The variable $x$ is absent, so this is Case 2. Substituting y' = p and y'' = p\dfrac{dp}{dy},

y p \frac{d p}{d y} + p^{2} = 0 .

Factor out $p$ :

p (y \frac{d p}{d y} + p) = 0 .

The factor $p = 0$ gives y' = 0, so $y = constant$ is a family of solutions. For $p \neq 0$ , separate the remaining factor:

y \frac{d p}{d y} = - p ⟹ \frac{d p}{p} = - \frac{d y}{y} ⟹ \ln | p | = - \ln | y | + const,

so $p = \frac{c_{1}}{y}$ .

Now replace $p$ by $\frac{d y}{d x}$ and separate again:

\frac{d y}{d x} = \frac{c_{1}}{y} ⟹ y d y = c_{1} d x ⟹ \frac{y^{2}}{2} = c_{1} x + c_{2} .

The general solution is $y^{2} = 2 c_{1} x + 2 c_{2}$ , or more simply $y^{2} = A x + B$ .

Solve y'' = -\dfrac{1}{y^{2}} with $y > 0$ , and interpret the result.

Solution

Since $x$ is absent, substitute y'' = p\dfrac{dp}{dy}:

p \frac{d p}{d y} = - \frac{1}{y^{2}} ⟹ p d p = - \frac{d y}{y^{2}} .

Integrating,

\frac{p^{2}}{2} = \frac{1}{y} + c_{1} ⟹ p = \pm \sqrt{\frac{2}{y} + 2 c_{1}} .

This relation is an energy integral. Reading $y$ as a position and $p$ as a velocity, it says that $\frac{1}{2} p^{2} - \frac{1}{y}$ stays constant along every solution, which is conservation of energy for an inverse-square attraction. Case 2 always produces a relation of this kind, because eliminating $x$ is precisely what a conservation law does.

The final step, $\frac{d y}{d x} = \pm \sqrt{\frac{2}{y} + 2 c_{1}}$ , is separable but its integral is not elementary for general $c_{1}$ .

Exercises

Solve y'' = 2x\left(y'\right)^{2}.

Solution

The variable $y$ is missing, so let p = y':

\frac{d p}{d x} = 2 x p^{2} ⟹ \frac{d p}{p^{2}} = 2 x d x ⟹ - \frac{1}{p} = x^{2} + c_{1} .

Therefore $p = \frac{- 1}{x^{2} + c_{1}}$ , and

y = - \int \frac{d x}{x^{2} + c_{1}} + c_{2} .

For $c_{1} = a^{2} > 0$ this gives $y = - \frac{1}{a} \arctan \frac{x}{a} + c_{2}$ . Note also the separate solution $p = 0$ , that is, $y = constant$ , lost when dividing by $p^{2}$ .

Solve y'' + \left(y'\right)^{2} = 0 by both cases, since neither $x$ nor $y$ appears.

Solution

As Case 1. With p = y', the equation is $\frac{d p}{d x} = - p^{2}$ , so $- \frac{1}{p} = - x + c$ , giving $p = \frac{1}{x - c}$ and

y = \ln | x - c | + c_{2} .

As Case 2. With y'' = p\dfrac{dp}{dy}, the equation is $p \frac{d p}{d y} = - p^{2}$ , so for $p \neq 0$ , $\frac{d p}{d y} = - p$ and $p = A e^{- y}$ . Then $\frac{d y}{d x} = A e^{- y}$ separates to $e^{y} d y = A d x$ , giving $e^{y} = A x + B$ and

y = \ln | A x + B |,

the same family after renaming constants.

Either route works when both variables are missing; pick whichever produces the simpler first-order equation.

Solve y'' = \left(y'\right)^{2}\tan y with the conditions $y (0) = 0$ and y'(0) = 1.

Solution

The variable $x$ is absent, so substitute y'' = p\dfrac{dp}{dy}:

p \frac{d p}{d y} = p^{2} \tan y ⟹ \frac{d p}{p} = \tan y d y (p \neq 0) .

Integrating, $\ln | p | = - \ln | \cos y | + const$ , so $p = \frac{A}{\cos y} = A \sec y$ .

The condition y'(0) = 1 with $y (0) = 0$ gives $1 = A \sec 0 = A$ , so $p = \sec y$ .

Now $\frac{d y}{d x} = \sec y$ separates as $\cos y d y = d x$ , giving $\sin y = x + B$ . Applying $y (0) = 0$ gives $B = 0$ , so

\sin y = x ⟹ y = \arcsin x,

valid on $- 1 < x < 1$ .

Frequently Asked Questions

When can I use reduction of order?

When the second-order equation is missing one of the two variables. If $y$ is absent, use p = y' with y'' = dp/dx and keep $x$ as the independent variable. If $x$ is absent, use p = y' with y'' = p\,dp/dy and switch to $y$ as the independent variable. If both $x$ and $y$ appear, this technique does not apply.

Why is y'' = p\,dp/dy in Case 2?

Because $p$ is being regarded as a function of $y$ rather than of $x$ . The chain rule gives

\frac{d p}{d x} = \frac{d p}{d y} \cdot \frac{d y}{d x} = \frac{d p}{d y} \cdot p .

Forgetting the extra factor of $p$ is the single most common mistake in Case 2 problems.

How many arbitrary constants should the answer have?

Two, since the equation is second order. One appears when you solve the reduced first-order equation for $p$ , and the second when you integrate $p$ to recover $y$ . If your answer has only one constant, you have almost certainly stopped a step early, at the formula for $p$ rather than for $y$ .

Is this the same "reduction of order" used for linear equations later?

The name is shared but the technique differs. Here, order is lowered by exploiting a missing variable. In the higher order chapter, reduction of order means using one known solution $y_{1}$ of a linear equation and substituting $y = v (x) y_{1} (x)$ to find a second one. Both lower the order by one, but they apply in different circumstances.

What does the intermediate relation between

p

and

y

mean physically?

In Case 2 it is usually a conservation law. Reading $y$ as position and $p$ as velocity, an equation such as $\frac{1}{2} p^{2} = - V (y) + c$ is conservation of energy: kinetic plus potential is constant along the motion. This is why Case 2 arises so often in mechanics, where the force depends on position but not explicitly on time.

Why does dividing by

p

deserve care?

Because $p = 0$ means y' = 0, that is, $y$ constant, and those constant solutions are genuine. Dividing by $p$ discards them, exactly as dividing by $y$ discarded equilibria in the separable and Bernoulli sections. Note the roots before dividing and restore them at the end.

Quick Reference

The Setting

Case 1: The Dependent Variable Does Not Appear

Case 2: The Independent Variable Does Not Appear

Exercises

Frequently Asked Questions

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