2.5 Reduction of Order

Turning a second-order equation with variable coefficients into a first-order linear equation.

What You'll Learn

By the end of this lesson, you will be able to:

Recognise when $y$ itself is missing ($c(x) = 0$) and reduce with $v = y'$
Solve the resulting first-order linear equation for $v$, then integrate to get $y$
Apply initial conditions in the right order: $y'$ on $v$, then $y$ on $y$
Use a known solution $y_c$ and the substitution $y = u\,y_c$ when $c(x) \neq 0$
Find a second solution, and a particular solution, from one known solution

Our methods for constant coefficients (2.1–2.3) and for Euler equations (2.4) do not reach a general equation $a(x)\,y'' + b(x)\,y' + c(x)\,y = f(x)$. Reduction of order handles two situations that come up often: either the $y$ term is missing, or one solution of the homogeneous equation is already known. In both, a clever substitution lowers the order from two to one, and Chapter 1 finishes the job.

Which Case Am I In?

The equation $a(x)\,y'' + b(x)\,y' + c(x)\,y = f(x)$ has…	Substitution	You get
Case 1: no $y$ term, $c(x) = 0$	$v = y'$, $v' = y''$	$a(x)\,v' + b(x)\,v = f(x)$, first-order linear in $v$
Case 2: $c(x) \neq 0$, and one solution $y_c$ of the homogeneous equation is given	$y = u\,y_c$, then $w = u'$	An equation with no $u$ term, so Case 1 in $u$
$c(x) \neq 0$ and no solution is known	Constant coefficients: 2.1. Euler form $ax^2y'' + bxy' + cy$: 2.4.

Case 1: The $y$ Term Is Missing, $c(x) = 0$

The Substitution

For an equation with no $y$ term,

a(x)\,y'' + b(x)\,y' = f(x),

let $v = y'$, so $v' = y''$. The equation becomes

a(x)\,v' + b(x)\,v = f(x),

a first-order linear equation in $v$, solved with an integrating factor (Chapter 1).

Check that $y$ does not appear. Only $y''$, $y'$ and functions of $x$ are allowed.
Substitute $v = y'$, $v' = y''$. Write the first-order equation in $v$.
Solve for $v$. Divide by $a(x)$ to get $v' + p(x)\,v = g(x)$, use $\mu = e^{\int p\,dx}$. You get one constant, $C_1$.
Apply a $y'$ condition now, if there is one. $y'(x_0)$ is $v(x_0)$.
Integrate $v$ to get $y$. $y = \int v\,dx + C_2$. This gives the second constant. Then apply any $y(x_0)$ condition.

Do Not Stop at $v$

$v$ is $y'$, not $y$. A second-order equation needs two constants: one from solving for $v$, one from integrating $v$.

Case 2: $c(x) \neq 0$, One Solution $y_c$ Is Known

If $y_c$ solves the homogeneous equation $a\,y'' + b\,y' + c\,y = 0$, look for solutions of the form $y = u(x)\,y_c(x)$. The product rule gives

y' = u'\,y_c + u\,y_c', \qquad y'' = u''\,y_c + 2u'\,y_c' + u\,y_c''.

Why the Order Drops

Substituting into $a\,y'' + b\,y' + c\,y = f$ and collecting terms:

a\,y_c\,u'' + \big(2a\,y_c' + b\,y_c\big)\,u' + \underbrace{\big(a\,y_c'' + b\,y_c' + c\,y_c\big)}_{=\,0 \text{ since } y_c \text{ is a solution}}\,u = f.

The $u$ term is gone, so this is Case 1 with $u$ in place of $y$. Let $w = u'$: $\;a\,y_c\,w' + \big(2a\,y_c' + b\,y_c\big)\,w = f$, a first-order linear equation.

Check $y_c$. Substitute it into the homogeneous equation and confirm you get 0.
Set $y = u\,y_c$. Compute $y'$ and $y''$ with the product rule.
Substitute and simplify. The $u$ term cancels. If it does not, $y_c$ or the algebra is wrong.
Let $w = u'$ and solve the first-order linear equation for $w$.
Integrate $w$ to get $u$, keeping both constants.
Form $y = u\,y_c$. The constants give $C_1 y_c + C_2 y_2$, plus $y_p$ when $f \neq 0$.

Shortcut for a Second Solution (Homogeneous Equation Only)

Written as $y'' + p(x)\,y' + q(x)\,y = 0$ (divide by $a(x)$ first): $\;\displaystyle y_2 = y_c \int \frac{e^{-\int p\,dx}}{y_c^{\,2}}\,dx.$ It is the result of the steps above; use it to check your work.

Common Mistakes

Using $v = y'$ when $y$ does appear. The $c(x)\,y$ term leaves two unknowns, $y$ and $v$.
Skipping the check that $y_c$ is a solution. If it is not, the $u$ term does not cancel.
Using the integrating factor without first dividing by the coefficient of $w'$.
Reporting $w$ or $u$ as the answer. The solution is $y = u\,y_c$.

Worked Examples

Examples 1–3 are Case 1 and Examples 4–8 are Case 2. Click each example to reveal the solution.

Spot the case. There is no $y$ term, so $c(x) = 0$.

Substitute $v = y'$. $x\,v' + v = 0$.

Solve for $v$. The left side is exactly $(x\,v)'$, so $(x\,v)' = 0$, $x\,v = C_1$, and $v = \dfrac{C_1}{x}$.

Integrate $v$. $y = \displaystyle\int \frac{C_1}{x}\,dx = C_1 \ln x + C_2$.

Solution

$$\boxed{y = C_1 \ln x + C_2}$$

Substitute $v = y'$. $x\,v' - v = 3x^2$.

Standard form. Divide by $x$: $v' - \dfrac{1}{x}\,v = 3x$.

Integrating factor. $\mu = e^{\int -1/x\,dx} = \dfrac{1}{x}$, so $\left(\dfrac{v}{x}\right)' = 3$.

Solve for $v$. $\dfrac{v}{x} = 3x + C_1$, so $v = 3x^2 + C_1 x$.

Integrate $v$. $y = x^3 + \dfrac{C_1}{2}x^2 + C_2$. Renaming $C_1/2$ as $C_1$:

Solution

$$\boxed{y = x^3 + C_1 x^2 + C_2}$$

Substitute $v = y'$. $x\,v' + 2v = 6x$.

Integrating factor. Multiplying by $x$ makes the left side a derivative: $x^2 v' + 2x\,v = (x^2 v)' = 6x^2$.

Solve for $v$. $x^2 v = 2x^3 + C_1$, so $v = 2x + \dfrac{C_1}{x^2}$.

Apply $y'(1) = 0$ to $v$. $v(1) = 2 + C_1 = 0$, so $C_1 = -2$ and $v = 2x - \dfrac{2}{x^2}$.

Integrate $v$. $y = x^2 + \dfrac{2}{x} + C_2$.

Apply $y(1) = 2$. $1 + 2 + C_2 = 2$, so $C_2 = -1$.

Solution

$$\boxed{y = x^2 + \frac{2}{x} - 1}$$

Check $y_c$. $y_c' = 2x$, $y_c'' = 2$: $x^2(2) - 3x(2x) + 4x^2 = 0$. ✓

Set $y = u\,x^2$. $y' = x^2 u' + 2x\,u$, $\;y'' = x^2 u'' + 4x\,u' + 2u$.

Substitute.

$$x^2(x^2u'' + 4xu' + 2u) - 3x(x^2u' + 2xu) + 4x^2u = x^4u'' + x^3u' + (2 - 6 + 4)x^2u$$

The $u$ term cancels, leaving $x^4 u'' + x^3 u' = 0$, or $x\,u'' + u' = 0$.

Recognise Case 1. This is Example 1 with $u$ in place of $y$, so $u = C_2 \ln x + C_1$.

Form $y = u\,x^2$.

Solution

$$\boxed{y = C_1 x^2 + C_2\,x^2 \ln x}$$

This is where the $\ln x$ in the repeated-root case of Euler equations (2.4) comes from.

Check $y_c$. $(x-1)e^x - x\,e^x + e^x = 0$. ✓

Set $y = u\,e^x$. $y' = (u' + u)\,e^x$, $\;y'' = (u'' + 2u' + u)\,e^x$.

Substitute and divide by $e^x$.

(x-1)(u'' + 2u' + u) - x(u' + u) + u = (x-1)\,u'' + (x-2)\,u' = 0

Let $w = u'$. $(x-1)\,w' + (x-2)\,w = 0$ is separable:

\frac{dw}{w} = -\frac{x-2}{x-1}\,dx = \left(-1 + \frac{1}{x-1}\right)dx

Solve for $w$. $\ln|w| = -x + \ln(x-1) + K$, so $w = C\,(x-1)\,e^{-x}$.

Integrate $w$. By parts, $\displaystyle\int (x-1)\,e^{-x}\,dx = -x\,e^{-x}$, so $u = -C\,x\,e^{-x} + D$.

Form $y = u\,e^x$. $y = -C\,x + D\,e^x$. The new solution is $y_2 = x$. (Check: $(x-1)\cdot 0 - x\cdot 1 + x = 0$.)

Solution

$$\boxed{y = C_1 e^x + C_2\,x}$$

Set $y = u\,e^{2x}$. $y' = (u' + 2u)\,e^{2x}$, $\;y'' = (u'' + 4u' + 4u)\,e^{2x}$.

Substitute and divide by $e^{2x}$.

$$(u'' + 4u' + 4u) - 4(u' + 2u) + 4u = u'' = 0$$

Solve. $u'' = 0$ gives $u = C_1 + C_2\,x$.

Solution

$$\boxed{y = (C_1 + C_2\,x)\,e^{2x}}$$

Reduction of order is the reason a repeated root $r$ in 2.1 gives $x\,e^{rx}$.

Check $y_c$ in the homogeneous equation. $x\,e^x - (x+1)\,e^x + e^x = 0$. ✓

Set $y = u\,e^x$ and substitute. With the derivatives from Example 5:

e^x\big[x\,u'' + (2x - x - 1)\,u' + (x - x - 1 + 1)\,u\big] = x^2 e^x \;\Longrightarrow\; x\,u'' + (x-1)\,u' = x^2

Let $w = u'$, standard form. $w' + \left(1 - \dfrac{1}{x}\right)w = x$.

Integrating factor. $\mu = e^{\int (1 - 1/x)\,dx} = \dfrac{e^x}{x}$, so $\left(\dfrac{e^x}{x}\,w\right)' = e^x$.

Solve for $w$. $\dfrac{e^x}{x}\,w = e^x + C$, so $w = x + C\,x\,e^{-x}$.

Integrate $w$. $u = \dfrac{x^2}{2} - C\,(x+1)\,e^{-x} + D$.

Form $y = u\,e^x$. $y = \dfrac{x^2}{2}e^x - C\,(x+1) + D\,e^x$.

Solution

$$\boxed{y = C_1 e^x + C_2\,(x+1) + \tfrac{1}{2}x^2 e^x}$$

One substitution produced both the second solution $y_2 = x + 1$ and the particular solution $y_p = \tfrac12 x^2 e^x$.

Check $y_c$. $x^2(0) + x(1) - x = 0$. ✓

Set $y = u\,x$. $y' = x\,u' + u$, $\;y'' = x\,u'' + 2u'$.

Substitute. $x^2(x\,u'' + 2u') + x(x\,u' + u) - u\,x = x^3 u'' + 3x^2 u' = x^2 + 1$.

Let $w = u'$. $x^3 w' + 3x^2 w = x^2 + 1$. The left side is already $(x^3 w)'$.

Solve for $w$. $x^3 w = \dfrac{x^3}{3} + x + C$, so $w = \dfrac13 + \dfrac{1}{x^2} + \dfrac{C}{x^3}$.

Integrate $w$. $u = \dfrac{x}{3} - \dfrac{1}{x} - \dfrac{C}{2x^2} + D$.

Form $y = u\,x$. $y = \dfrac{x^2}{3} - 1 - \dfrac{C}{2x} + D\,x$.

Solution

$$\boxed{y = C_1 x + \frac{C_2}{x} + \frac{x^2}{3} - 1}$$

📝 Exam-Style Practice Problems

Try each problem first, then click it to check your solution.

Practice 1: Solve $(1+x^2)\,y'' + 2x\,y' = 0$, $y(0) = 1$, $y'(0) = 2$ ▶

Solution

No $y$ term, so let $v = y'$: $(1+x^2)\,v' + 2x\,v = 0$, which is $\big((1+x^2)\,v\big)' = 0$.

So $v = \dfrac{C_1}{1+x^2}$. From $y'(0) = 2$: $C_1 = 2$.

Integrate: $y = 2\arctan x + C_2$. From $y(0) = 1$: $C_2 = 1$.

\boxed{y = 1 + 2\arctan x}

Practice 2: Solve $y'' + 2y' = 4x$ using $v = y'$ ▶

Solution

$v' + 2v = 4x$. Integrating factor $e^{2x}$: $\left(e^{2x}v\right)' = 4x\,e^{2x}$.

By parts, $e^{2x}v = (2x - 1)\,e^{2x} + C$, so $v = 2x - 1 + C\,e^{-2x}$.

Integrate, renaming $-C/2$ as $C_1$:

\boxed{y = x^2 - x + C_1 e^{-2x} + C_2}

Methods 2.1 and 2.2 give the same answer. With $c = 0$, reduction of order is often quicker.

Practice 3: $x^2 y'' + x\,y' - 4y = 0$ has the solution $y_c = x^2$. Find the general solution. ▶

Solution

Set $y = u\,x^2$: $y' = x^2u' + 2xu$, $y'' = x^2u'' + 4xu' + 2u$.

Substituting, the $u$ term cancels: $x^4 u'' + 5x^3 u' = 0$, so $x\,w' + 5w = 0$ with $w = u'$.

Separable: $w = C\,x^{-5}$, so $u = -\dfrac{C}{4x^4} + D$ and $y_2 \propto x^2 \cdot x^{-4} = x^{-2}$.

\boxed{y = C_1 x^2 + \frac{C_2}{x^2}}

Practice 4: $x^2 y'' - 2x\,y' + 2y = x^3$, $x > 0$, with $y_c = x$. Find the general solution. ▶

Solution

Set $y = u\,x$: $y' = xu' + u$, $y'' = xu'' + 2u'$.

Substitute: $x^2(xu'' + 2u') - 2x(xu' + u) + 2ux = x^3 u''$. Both the $u$ and $u'$ terms cancel, so $x^3u'' = x^3$ and $u'' = 1$.

Integrate twice: $u = \dfrac{x^2}{2} + C_2\,x + C_1$, and $y = u\,x$:

\boxed{y = C_1 x + C_2 x^2 + \frac{x^3}{2}}

Practice Problems

Test your understanding of reduction of order. Choosing an answer shows why it is right or wrong.

Progress: 0 / 8 completed

The equation \(a(x)\,y'' + b(x)\,y' + c(x)\,y = f(x)\) has…	Substitution	You get
Case 1: no \(y\) term, \(c(x) = 0\)	\(v = y'\), \(v' = y''\)	\(a(x)\,v' + b(x)\,v = f(x)\), first-order linear in \(v\)
Case 2: \(c(x) \neq 0\), and one solution \(y_c\) of the homogeneous equation is given	\(y = u\,y_c\), then \(w = u'\)	An equation with no \(u\) term, so Case 1 in \(u\)
\(c(x) \neq 0\) and no solution is known	Constant coefficients: 2.1. Euler form \(ax^2y'' + bxy' + cy\): 2.4.

2.5 Reduction of Order

Which Case Am I In?

Case 1: The \(y\) Term Is Missing, \(c(x) = 0\)

Case 2: \(c(x) \neq 0\), One Solution \(y_c\) Is Known

Worked Examples

📝 Exam-Style Practice Problems

Solution

Solution

Solution

Solution

Practice Problems

Q1: Which substitution turns $x\,y'' + 3y' = x$ into a first-order linear equation?

Q2: With $v = y'$, the equation $x\,y'' + y' = 4x$ becomes:

Q3: The general solution of $x\,y'' + y' = 4x$, $x > 0$, is:

Q4: When does $v = y'$ reduce $a(x)\,y'' + b(x)\,y' + c(x)\,y = f(x)$ to a first-order linear equation in $v$?

Q5: Substituting $y = u\,y_c$ into $a\,y'' + b\,y' + c\,y = f$, why does the term in $u$ (no derivative) disappear?

Q6: $x^2 y'' + x\,y' - y = 0$ has the solution $y_c = x$. Reduction of order gives a second solution:

Q7: For $y'' - 6y' + 9y = 0$ with $y_c = e^{3x}$, setting $y = u\,e^{3x}$ gives:

Q8: For $x\,y'' - (x+1)\,y' + y = 0$ with $y_c = e^x$, which equation does $w = u'$ satisfy?