Calculus II · Unit 4B · lesson

A Preview of Series Solutions to Differential Equations

Substitute a power series into a differential equation, match coefficients, and derive a recurrence for the solution.

Core textbook pathFollow the sequence or choose a destination from the unit map.Open the Unit 4B map →

Section overview

Taylor bounds and approximation

What this section is building

Substitute a power series into a differential equation, match coefficients, and derive a recurrence for the solution.

Notice

The polynomial supplies the estimate; the remainder theorem supplies the trust boundary.

Decide

Choose a tractable center and degree, bound the needed derivative on the whole interval, then compare the bound with the required tolerance.

Avoid

Evaluating the next term without checking that the theorem's hypotheses make it a valid error bound.

Concept

Learning objectives

use coefficient matching to see how a differential equation can determine a power series recursively.

A Preview of Series Solutions to Differential Equations

Explanation

A differential equation can solve for coefficients one at a time

Suppose a solution is represented by $y=\sum c_nx^n$ . Differentiating the series and substituting into a differential equation turns the equation into an identity between power series. Equal power series have equal coefficients, so the differential equation generates recurrence relations among the $c_n$ .

This method extends beyond equations with elementary closed forms and leads to special functions. In this unit we use it only as a preview, because a full treatment belongs with differential equations. The important idea is that an infinite function problem becomes a systematic algebra problem in coefficients.

Bridge

An unknown function can be replaced by unknown coefficients

To solve a differential equation by power series, assume the solution has the form $y=\sum a_nx^n$ . Differentiate term by term, substitute the series into the equation, align powers, and compare coefficients. The differential equation becomes a recurrence for the numbers $a_n$ .

This method extends polynomial reasoning rather than introducing magic. Initial conditions determine starting coefficients, and the recurrence determines the rest. A full theory must also prove convergence, but the coefficient mechanism already previews why power series are central in differential equations and mathematical physics.

**A differential equation becomes a recurrence**
A flow diagram starts with the initial coefficient a0 and repeatedly applies the next-coefficient recurrence, producing factorial denominators in succession.
A differential equation becomes a recurrence. A coefficient machine mapping a n to a n+1 .

Proof idea

Equal power series have equal coefficients

Inside a common convergence interval, if two power series represent the same function, differentiating repeatedly at the center isolates each coefficient. This uniqueness lets the differential equation determine a recurrence term by term.

Concept

Coefficient matching

If two power series agree on an interval,

\sum a_nx^n=\sum b_nx^n,

then $a_n=b_n$ for every $n$ .

Guided walkthrough

Let $y=\sum_{n=0}^{\infty}c_nx^n$ . Then

y'=\sum_{n=0}^{\infty}(n+1)c_{n+1}x^n.

The equation $y'=y$ gives $(n+1)c_{n+1}=c_n$ , so $c_n=c_0/n!$ . Therefore

y=c_0\sum_{n=0}^{\infty}\frac{x^n}{n!}=c_0e^x.

Worked example

Recover the exponential from y prime equals y

Assume

y=\sum_{n=0}^{\infty}a_nx^n.

Then

y'=\sum_{n=0}^{\infty}(n+1)a_{n+1}x^n.

The equation $y'=y$ gives

(n+1)a_{n+1}=a_n, \qquad a_{n+1}=\frac{a_n}{n+1}.

If $y(0)=1$ , then $a_0=1$ , so $a_n=1/n!$ and

y=\sum_{n=0}^{\infty}\frac{x^n}{n!}=e^x.

Common mistake

Align the index before comparing coefficients

The series for $y'$ initially begins with $x^{n-1}$ . Reindex it so both sides use the same power $x^n$ before equating coefficients.

Interactive checku4b-series_solutions_preview-01

If $(n+1)c_{n+1}=c_n$ , express $c_n$ in terms of $c_0$ .

Your work stays on this device. No account or AI grader is used.

Show hint

Compute the first few coefficients.

Attempt once to unlock the solution

Submit an answer first. The hint is available now.

Exercise

Use a series to analyze $y^{\prime\prime}+y=0$ .

Exercise

Explain how initial conditions determine free coefficients.

Exercise

Why must indices be shifted before coefficient matching?

Exercise

Name one reason series solutions are useful.

Source & rights

Original instruction with traceable references.

BetterGrades-original; no direct adaptation declared in the verified handoff.

Reference textbooks remain rights-separated and are not published as application assets. Any direct adaptation requires separate identification and attribution.

A Preview of Series Solutions to Differential Equations

Explanation

A differential equation can solve for coefficients one at a time

Bridge

An unknown function can be replaced by unknown coefficients

Read this graph as text

A differential equation becomes a recurrence. A flow diagram starts with the initial coefficient a0 and repeatedly applies a n+1 A coefficient machine mapping a n to a n+1 . Preserve the misconception control described in the lesson and storyboard.

Written labels, distinct line styles, markers, and fill patterns communicate every relationship in a differential equation becomes a recurrence; color is never the only cue.

Why it matters: A coefficient machine mapping a n to a n+1 .

A differential equation becomes a recurrence

A flow diagram starts with the initial coefficient a0 and repeatedly applies the next-coefficient recurrence, producing factorial denominators in succession.

A differential equation becomes a recurrence. A coefficient machine mapping a n to a n+1 .

Proof idea

Equal power series have equal coefficients

Concept

Coefficient matching

If two power series agree on an interval,

\sum a_nx^n=\sum b_nx^n,

then $a_n=b_n$ for every $n$ .

Guided walkthrough

Let $y=\sum_{n=0}^{\infty}c_nx^n$ . Then

y'=\sum_{n=0}^{\infty}(n+1)c_{n+1}x^n.

The equation $y'=y$ gives $(n+1)c_{n+1}=c_n$ , so $c_n=c_0/n!$ . Therefore

y=c_0\sum_{n=0}^{\infty}\frac{x^n}{n!}=c_0e^x.

Worked example

Recover the exponential from y prime equals y

Assume

y=\sum_{n=0}^{\infty}a_nx^n.

Then

y'=\sum_{n=0}^{\infty}(n+1)a_{n+1}x^n.

The equation $y'=y$ gives

(n+1)a_{n+1}=a_n, \qquad a_{n+1}=\frac{a_n}{n+1}.

If $y(0)=1$ , then $a_0=1$ , so $a_n=1/n!$ and

y=\sum_{n=0}^{\infty}\frac{x^n}{n!}=e^x.

Common mistake

Align the index before comparing coefficients

The series for $y'$ initially begins with $x^{n-1}$ . Reindex it so both sides use the same power $x^n$ before equating coefficients.

Interactive checku4b-series_solutions_preview-01

If $(n+1)c_{n+1}=c_n$ , express $c_n$ in terms of $c_0$ .

Your work stays on this device. No account or AI grader is used.

Show hint

Compute the first few coefficients.

Attempt once to unlock the solution

Submit an answer first. The hint is available now.

Exercise

Use a series to analyze $y^{\prime\prime}+y=0$ .

Exercise

Explain how initial conditions determine free coefficients.

Exercise

Why must indices be shifted before coefficient matching?

Exercise

Name one reason series solutions are useful.

Source & rights

Original instruction with traceable references.

BetterGrades-original; no direct adaptation declared in the verified handoff.

Reference textbooks remain rights-separated and are not published as application assets. Any direct adaptation requires separate identification and attribution.

A Preview of Series Solutions to Differential Equations

Related resources

Original instruction with traceable references.

A Preview of Series Solutions to Differential Equations

Related resources

Original instruction with traceable references.