Working out the coefficients · Differential Equations

Intuition

With variable coefficients the method is the same, and now it earns its keep. Each term of the equation contributes a sum; after shifting, the coefficient of each power gives one equation, and solving it for the coefficient with the highest index gives that coefficient from earlier ones. The coefficients split into two families, one started by the initial value and one by the initial slope, and each family is one solution.

A row of dominoes. Once the first two are pushed — the initial value and the initial slope — the equation knocks down every later one in turn.

plane diagram: a-function-the-method-found

$y''-xy'-y=0$ with $y(0)=1$ , $y'(0)=0$ . The recurrence $a_{n+2}=\tfrac{a_{n}}{n+2}$ gives $1+\tfrac{x^{2}}{2}+\tfrac{x^{4}}{8}+\cdots$ , which sums to $e^{x^{2}/2}$ : a solution the method found without it being guessed.

Two families of coefficients

For $y''-xy'-y=0$ , substituting gives $\sum [$ , so . The recurrence steps by two, so the even coefficients follow from and the odd ones from . The general solution is , where is the even series started at and the odd series started at .

a_{n+2}=\frac{a_{n}}{n+2},\qquad n=0,1,2,\ldots

The two families

Even: $a_{2}=\tfrac{a_{0}}{2}$ , , , so .

The even and odd series are a fundamental pair

\begin{aligned}&y_{1}(0)=1,\ y_{1}'(0)=0,\ y_{2}(0)=0,\ y_{2}'(0)=1\\&\Longrightarrow\ W(0)=1\ne0\end{aligned}

The even series starts with 1 and has no x term, and the odd series starts with x. So at zero the first has value one and slope zero, and the second value zero and slope one. Their Wronskian there is one, so by the theorem of chapter four they form a fundamental pair, and every solution is a combination of them with the initial value and the initial slope as the constants.

Proof steps

y_{1}=1+a_{2}x^{2}+\cdots,\qquad y_{2}=x+a_{3}x^{3}+\cdots

The even series starts at 1 and the odd one at x.

y_{1}(0)=1,\ y_{1}'(0)=0,\ y_{2}(0)=0,\ y_{2}'(0)=1

Read the values and slopes at zero off the first terms.

W(0)=1\cdot1-0\cdot0=1

The Wronskian at zero is not zero.

y=a_{0}\,y_{1}+a_{1}\,y_{2}

So every solution is a combination of the two, by the theorem of the Wronskian lesson.

Applications

Practice

From the Equation to a Recurrence

Substitute, shift, collect the coefficient of xⁿ and solve it for the coefficient with the highest index.

y''-xy'-y=0\ \Longrightarrow\ a_{n+2}=\frac{a_{n}}{n+2}

Try it

Which recurrence does $y''-y=0$ give?