Question

use power series to solve the initial value problem.

$y^{primeprime}+xy^{prime}-2y = 0$; $y(0)=3$, $y^{prime}(0)=0$

$y(x)=\square$

Explanation:

Step1: Assume power series solution

Assume \(y=\sum_{n = 0}^{\infty}a_{n}x^{n}\), then \(y'=\sum_{n = 1}^{\infty}na_{n}x^{n - 1}\) and \(y''=\sum_{n = 2}^{\infty}n(n - 1)a_{n}x^{n - 2}\).

Step2: Substitute into the differential equation

Substitute \(y\), \(y'\), and \(y''\) into \(y''+xy'-2y = 0\).

• For \(y''\): \(\sum_{n = 2}^{\infty}n(n - 1)a_{n}x^{n - 2}=\sum_{k=0}^{\infty}(k + 2)(k+1)a_{k + 2}x^{k}\) (let \(k=n - 2\)).
• For \(xy'\): \(x\sum_{n = 1}^{\infty}na_{n}x^{n - 1}=\sum_{n = 1}^{\infty}na_{n}x^{n}\).
• For \(-2y\): \(-2\sum_{n = 0}^{\infty}a_{n}x^{n}\).

The equation becomes \(\sum_{k = 0}^{\infty}(k + 2)(k + 1)a_{k+2}x^{k}+\sum_{n = 1}^{\infty}na_{n}x^{n}-2\sum_{n = 0}^{\infty}a_{n}x^{n}=0\).

Separate the \(k = 0\) term from the first - sum:
\(2a_{2}-2a_{0}+\sum_{k = 1}^{\infty}[(k + 2)(k + 1)a_{k+2}+ka_{k}-2a_{k}]x^{k}=0\).

Step3: Use initial conditions

Given \(y(0)=3\), so \(a_{0}=3\). Given \(y'(0) = 0\), and \(y'=\sum_{n = 1}^{\infty}na_{n}x^{n - 1}\), then \(a_{1}=0\).

From \(2a_{2}-2a_{0}=0\), substituting \(a_{0}=3\), we get \(a_{2}=3\).

For \(k\geq1\), the recurrence relation is \((k + 2)(k + 1)a_{k+2}+(k - 2)a_{k}=0\), so \(a_{k+2}=\frac{2 - k}{(k + 2)(k + 1)}a_{k}\).

Since \(a_{1}=0\), then \(a_{3}=a_{5}=\cdots=0\) (odd - indexed coefficients are zero).

For \(k = 2\): \(a_{4}=\frac{2-2}{4\times3}a_{2}=0\).

For \(k = 4\): \(a_{6}=\frac{2 - 4}{6\times5}a_{4}=0\), and in general, for \(k\geq2\) even (\(k = 2m,m\geq1\)), \(a_{2m + 2}=0\).

Step4: Write the power - series solution

\(y=a_{0}+a_{1}x+a_{2}x^{2}+a_{3}x^{3}+\cdots\). Substituting \(a_{0}=3\), \(a_{1}=0\), \(a_{2}=3\), \(a_{3}=0\), \(a_{4}=0,\cdots\)

\(y = 3+3x^{2}\).

Answer:

\(y(x)=3 + 3x^{2}\)