Question

find the first three nonzero terms in each of two linearly independent solutions of the form formula substitute known taylor series for the analytic functions and retain enough terms to compute the necessary coefficients

$y+e^{-x}y = 0$

enter the first three nonzero terms in each of two linearly independent solutions the first term of $y_1$ is given

$y_1(x)=1+\square+\cdots$

$y_2(x)=\square+\cdots$

Explanation:

Step1: Recall the Taylor series of \(e^{-x}\)

The Taylor series of \(e^{-x}=\sum_{n = 0}^{\infty}\frac{(-x)^{n}}{n!}=1 - x+\frac{x^{2}}{2!}-\frac{x^{3}}{3!}+\cdots\)

Let \(y=\sum_{n=0}^{\infty}a_{n}x^{n}=a_{0}+a_{1}x + a_{2}x^{2}+a_{3}x^{3}+\cdots\), then \(y''=\sum_{n = 2}^{\infty}n(n - 1)a_{n}x^{n-2}=\sum_{n=0}^{\infty}(n + 2)(n+1)a_{n+2}x^{n}\)

Substitute \(y\) and \(e^{-x}\) into the differential equation \(y''+e^{-x}y = 0\):

\(\sum_{n=0}^{\infty}(n + 2)(n + 1)a_{n+2}x^{n}+(1 - x+\frac{x^{2}}{2}-\frac{x^{3}}{6}+\cdots)(a_{0}+a_{1}x + a_{2}x^{2}+a_{3}x^{3}+\cdots)=0\)

For \(n = 0\):

\(2a_{2}+a_{0}=0\Rightarrow a_{2}=-\frac{a_{0}}{2}\)

For \(n = 1\):

\(6a_{3}+a_{1}-a_{0}=0\Rightarrow a_{3}=\frac{a_{0}-a_{1}}{6}\)

For \(n = 2\):

\(12a_{4}+a_{2}-a_{1}+\frac{a_{0}}{2}=0\)

Substitute \(a_{2}=-\frac{a_{0}}{2}\) into the above - equation:

\(12a_{4}-\frac{a_{0}}{2}-a_{1}+\frac{a_{0}}{2}=0\Rightarrow a_{4}=\frac{a_{1}}{12}\)

Step2: Find \(y_{1}(x)\) (let \(a_{0}=1,a_{1}=0\))

If \(a_{0} = 1,a_{1}=0\)

\(a_{2}=-\frac{1}{2},a_{3}=\frac{1}{6},a_{4}=0\)

\(y_{1}(x)=1-\frac{1}{2}x^{2}+\frac{1}{6}x^{3}+\cdots\)

Step3: Find \(y_{2}(x)\) (let \(a_{0}=0,a_{1}=1\))

If \(a_{0}=0,a_{1}=1\)

\(a_{2}=0,a_{3}=-\frac{1}{6},a_{4}=\frac{1}{12}\)

\(y_{2}(x)=x-\frac{1}{6}x^{3}+\frac{1}{12}x^{4}+\cdots\)

Answer:

\(y_{1}(x)=1-\frac{1}{2}x^{2}+\frac{1}{6}x^{3}+\cdots\)

\(y_{2}(x)=x-\frac{1}{6}x^{3}+\frac{1}{12}x^{4}+\cdots\)