Q9 E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Find at least the first four non-zero terms in a power series expansion about x₀ for a general solution to the given differential equation with the value for x₀.

$\left(x^2-2x\right)\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{2}\mathbf{y}\mathbf{=}\mathbf{0}\mathbf{;}\mathbf{ }{\mathbf{x}}_{\mathbf{0}}\mathbf{=}\mathbf{1}$

Step-by-Step Solution

Answer

The first four nonzero terms in a power series expansion about x₀ for a general solution:

$\mathrm{y}={\mathrm{a}}_0\left(1+{(\mathrm{x}+1)}^2+\frac{1}{3}{(\mathrm{x}-1)}^4+...\right)+{\mathrm{a}}_1\left(\left(\mathrm{x}-1\right)+\frac{1}{3}{(\mathrm{x}-1)}^3+...\right)$

Step 1: Power series expansion

A power series expansion of can be obtained simply by expanding the exponential and integrating term-by-term. This series converges for all, but the convergence becomes extremely slow if significantly exceeds unity.

Step 2: To determine the first four nonzero terms in a power series expansion about x 0 for a general solution.

$\begin{array}{l}\left({\mathrm{x}}^2-2\mathrm{x}\right)\mathrm{y}\text{'}\text{'}+2\mathrm{y}=0\\ \mathrm{y}\text{'}\text{'}+\frac{2}{{\mathrm{x}}^2-2\mathrm{x}}\mathrm{y}=0\end{array}$

We get the value of $\mathrm{q}\left(\mathrm{x}\right)=\frac{2}{{\mathrm{x}}^2-2\mathrm{x}}$.

$\begin{array}{l}\mathrm{y}\left(\mathrm{x}\right)=\sum _{\mathrm{n}=0}^{\infty }{\mathrm{a}}_{\mathrm{n}}{\left(\mathrm{x}-1\right)}^{\mathrm{n}}\\ \mathrm{y}\text{'}\text{'}\left(\mathrm{x}\right)=\sum _{\mathrm{n}=2}^{\infty }\mathrm{n}(\mathrm{n}-1){\mathrm{a}}_{\mathrm{n}}{\left(\mathrm{x}-1\right)}^{\mathrm{n}-2}+2\sum _{\mathrm{n}=0}^{\infty }{\mathrm{a}}_{\mathrm{n}}{\left(\mathrm{x}-1\right)}^{\mathrm{n}}=0\end{array}$

Let, $\mathrm{x}-1=\mathrm{t}$:

$\begin{array}{l}(\mathrm{t}+1)(\mathrm{t}-1)\sum _{\mathrm{n}=2}^{\infty }\mathrm{n}(\mathrm{n}-1){\mathrm{a}}_{\mathrm{n}}(\mathrm{t}{)}^{\mathrm{n}-2}+2\sum _{\mathrm{n}=0}^{\infty }{\mathrm{a}}_{\mathrm{n}}{\left(\mathrm{t}\right)}^{\mathrm{n}}=0\\ \sum _{\mathrm{n}=2}^{\infty }\mathrm{n}(\mathrm{n}-1){\mathrm{a}}_{\mathrm{n}}(\mathrm{t}{)}^{\mathrm{n}}-\sum _{\mathrm{n}=2}^{\infty }\mathrm{n}(\mathrm{n}-1){\mathrm{a}}_{\mathrm{n}}(\mathrm{t}{)}^{\mathrm{n}-2}+2\sum _{\mathrm{n}=0}^{\infty }{\mathrm{a}}_{\mathrm{n}}{\left(\mathrm{t}\right)}^{\mathrm{n}}=0\\ 2{\mathrm{a}}_2{\mathrm{t}}^2+6{\mathrm{a}}_3{\mathrm{t}}^3+12{\mathrm{a}}_4{\mathrm{t}}^4+20{\mathrm{a}}_5{\mathrm{t}}^5+...\\ 2{\mathrm{a}}_0+2{\mathrm{a}}_1\mathrm{t}+2{\mathrm{a}}_2{\mathrm{t}}^2+2{\mathrm{a}}_3{\mathrm{t}}^3+2{\mathrm{a}}_4{\mathrm{t}}^4+...=0\\ (2{\mathrm{a}}_0-2{\mathrm{a}}_2)+(2{\mathrm{a}}_1-6{\mathrm{a}}_3)(\mathrm{x}-1)+(2{\mathrm{a}}_2-12{\mathrm{a}}_4+2{\mathrm{a}}_2)(\mathrm{x}-1{)}^2+(2{\mathrm{a}}_3-20{\mathrm{a}}_5)(\mathrm{x}-1{)}^3+....=0\\ 2{\mathrm{a}}_0-2{\mathrm{a}}_2=0\Rightarrow {\mathrm{a}}_2={\mathrm{a}}_0\\ 2{\mathrm{a}}_1-6{\mathrm{a}}_3=0\Rightarrow {\mathrm{a}}_3=\frac{1}{3}{\mathrm{a}}_1\\ 2{\mathrm{a}}_2-12{\mathrm{a}}_4+2{\mathrm{a}}_2=0\Rightarrow {\mathrm{a}}_4=\frac{1}{3}{\mathrm{a}}_0\end{array}$

In the end,

$\begin{array}{l}\mathrm{y}=\sum _{\mathrm{n}=0}^{\infty }{\mathrm{a}}_{\mathrm{n}}{\left(\mathrm{x}-1\right)}^{\mathrm{n}}={\mathrm{a}}_0+{\mathrm{a}}_1(\mathrm{x}-1)+{\mathrm{a}}_2(\mathrm{x}-1{)}^2+{\mathrm{a}}_3(\mathrm{x}-1{)}_3+...\\ \mathrm{y}={\mathrm{a}}_0\left(1+{(\mathrm{x}+1)}^2+\frac{1}{3}{(\mathrm{x}-1)}^4+...\right)+{\mathrm{a}}_1\left(\left(\mathrm{x}-1\right)+\frac{1}{3}{(\mathrm{x}-1)}^3+...\right)\end{array}$

Hence, the final answer is $\mathrm{y}={\mathrm{a}}_0\left(1+{(\mathrm{x}+1)}^2+\frac{1}{3}{(\mathrm{x}-1)}^4+...\right)+{\mathrm{a}}_1\left(\left(\mathrm{x}-1\right)+\frac{1}{3}{(\mathrm{x}-1)}^3+...\right)$