Reduction of Order

When reduction of order is the right tool

If you already know one solution $y_1(x)$ of a second-order linear homogeneous ODE, reduction of order lets you find a second, linearly independent solution $y_2(x)$ without guessing. It works for any second-order linear homogeneous ODE (constant or variable coefficients), making it more general than the characteristic-equation trick (#181–#185) which only handles constant coefficients.

The key idea: guess $y_2(x) = u(x) \, y_1(x)$ with $u$ non-constant and plug in. The ODE collapses from a 2nd-order equation on $y$ to a 1st-order equation on $u'$ — hence the name, "reduction of order."

This also provides the cleanest derivation of why $x e^{r x}$ is the second solution at a repeated root (#185).

The given problem

$y'' - 2 y' + y = 0$

Given that $y_1 = e^{x}$ solves this. (You can verify: $y_1'' - 2 y_1' + y_1 = e^{x} - 2 e^{x} + e^{x} = 0$ ✓. The characteristic equation is $(r - 1)^{2} = 0$, so this is a repeated-root case with $r = 1$.)

Step-by-step recipe

Step 1 — Ansatz.

Assume $y_2(x) = u(x) \, y_1(x) = u(x) \, e^{x}$ for some unknown function $u(x)$.

Step 2 — Compute derivatives (product rule). $y_2' = u' e^{x} + u \, e^{x} = (u' + u) e^{x}$ $y_2'' = (u'' + u') e^{x} + (u' + u) e^{x} = (u'' + 2 u' + u) e^{x}$

Step 3 — Substitute into the ODE. $y_2'' - 2 y_2' + y_2 = e^{x}\bigl[(u'' + 2 u' + u) - 2(u' + u) + u\bigr]$ $= e^{x}\bigl[u'' + 2 u' + u - 2 u' - 2 u + u\bigr] = e^{x} \cdot u''.$

Setting this to zero: $u'' = 0.$

Integrate twice: $u(x) = A + B x,$ where $A, B$ are integration constants.

Step 4 — Assemble. $y_2(x) = u(x) \, e^{x} = (A + B x) \, e^{x}.$

The constant term $A \, e^{x}$ is just a scalar multiple of $y_1 = e^{x}$ — it's not linearly independent. Throw it away (set $A = 0$) and take the non-constant part $B \, x \, e^{x}$. Pick $B = 1$: $\boxed{\, y_2(x) = x \, e^{x} \,}$

Verification

$y_2 = x e^{x}$, $y_2' = e^{x} + x e^{x} = (1 + x) e^{x}$, $y_2'' = e^{x} + (1 + x) e^{x} = (2 + x) e^{x}$.

Plug: $y_2'' - 2 y_2' + y_2 = (2 + x) e^{x} - 2 (1 + x) e^{x} + x e^{x}$ $= e^{x} \bigl[(2 + x) - 2(1 + x) + x\bigr] = e^{x} \bigl[2 + x - 2 - 2 x + x\bigr] = 0. \; \checkmark$

General solution

With both independent solutions in hand: $y(x) = C_1 \, e^{x} + C_2 \, x \, e^{x} = (C_1 + C_2 x) \, e^{x}.$

Exactly the repeated-root form we claimed in #185, now derived rather than pulled from a table.

Why it always works — the structural reason

For a 2nd-order linear homogeneous ODE $y'' + p(x) y' + q(x) y = 0$ with known solution $y_1$, plug in $y_2 = u \, y_1$ and simplify. The coefficient of $u$ is $y_1'' + p y_1' + q y_1 = 0$ (since $y_1$ solves the ODE), which kills the $u$-term. What survives is a 2nd-order linear ODE in $u$ with no $u$ term — equivalent to a 1st-order ODE in $u'$: $y_1 \, u'' + (2 y_1' + p y_1) \, u' = 0.$

Divide by $y_1$ (where $y_1 \ne 0$): $u'' + \left( 2 \frac{y_1'}{y_1} + p \right) u' = 0.$

Substitute $w = u'$: $w' + \left( 2 \frac{y_1'}{y_1} + p \right) w = 0.$

This is a first-order linear ODE in $w$, solved by the integrating-factor formula: $w(x) = u'(x) = \frac{C}{y_1^{2}(x)} \exp\!\left( -\int p(x) \, dx \right).$

Abel's formula emerges! Integrating $u'$ gives $u$, and $y_2 = u \, y_1$.

In our problem $p(x) = -2$, $y_1 = e^{x}$, so $y_1^{2} = e^{2x}$ and $\int p \, dx = -2 x$: $u'(x) = \frac{C}{e^{2x}} e^{2x} = C \implies u(x) = C x + D.$

Giving $y_2 = (C x + D) e^{x}$ — same answer, via the general formula.

Variable-coefficient example

$x^{2} y'' + x y' - y = 0, \quad y_1 = x.$

Cauchy-Euler (see #188). Indicial $r^{2} - 1 = 0 \implies r = \pm 1$, second solution $y_2 = 1/x$ by inspection. Let's get it by reduction of order instead.

$y_2 = u(x) \cdot x$. Then $y_2' = u' x + u$, $y_2'' = u'' x + 2 u'$. Substitute: $x^{2} (u'' x + 2 u') + x (u' x + u) - u x = x^{3} u'' + 2 x^{2} u' + x^{2} u' + x u - u x = x^{3} u'' + 3 x^{2} u'.$

Set to zero and divide by $x^{2}$: $x u'' + 3 u' = 0$, i.e. $u'' = -(3/x) u'$. Let $w = u'$: $w'/w = -3/x \implies \ln|w| = -3 \ln|x| + C \implies w = C/x^{3}$. Integrate: $u = -C/(2 x^{2}) + D$. Take $u = 1/x^{2}$ (dropping constants): $y_2 = u \cdot x = \frac{1}{x}. \; \checkmark$

Comparison to the limit derivation (for repeated roots)

In #185 we also derived $x e^{r x}$ as a limit of the distinct-roots formula: $\frac{e^{(r + \varepsilon) x} - e^{r x}}{\varepsilon} \xrightarrow{\varepsilon \to 0} x \, e^{r x}.$

Reduction of order works without needing to perturb and take a limit — it's a direct algorithm. Both perspectives produce the same answer, but reduction of order is the preferred method in practice because it:

1. Works for any known $y_1$, not just exponentials.
2. Works for variable-coefficient ODEs.
3. Doesn't require the perturbation setup.

Common mistakes

• Trying $y_2 = u \cdot y_1$ with $u$ constant. That just gives another multiple of $y_1$. The non-trivial content lives in the non-constant part of $u$.
• Forgetting to discard the redundant constant. After solving for $u$, the part of $u$ that's a constant times $y_1 / y_1 = 1$ just produces a multiple of $y_1$; drop it.
• Algebra errors in the product rule. $y_2''$ has three terms from the second derivative of a product — plus $u'' y_1$, $2 u' y_1'$, and $u y_1''$ — and they combine with the ODE's $p y_2'$ and $q y_2$ to produce the simplification. Track each piece.
• Dividing by $y_1$ near its zeros. If $y_1$ has zeros inside your interval, the derivation breaks there and you need to work on connected sub-intervals where $y_1 \ne 0$.

Try it in the visualization

Slide a "perturbation" parameter that moves the two roots of the characteristic equation apart. At the merged position, watch how the usual distinct-root solution bifurcates into the $\{e^{r x}, x e^{r x}\}$ basis — reduction of order shown as a continuous limit.