Q10 E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

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

${\mathbf{x}}^{\mathbf{2}}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{xy}\mathbf{\text{'}}\mathbf{+}\mathbf{2}\mathbf{y}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{ }{\mathbf{x}}_{\mathbf{0}}\mathbf{=}\mathbf{2}$

Step-by-Step Solution

Answer

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

$\mathrm{y}={\mathrm{a}}_0\left(1+\frac{1}{4}{(\mathrm{x}-2)}^2+\frac{1}{24}{(\mathrm{x}-2)}^3+...\right)+{\mathrm{a}}_1\left(\left(\mathrm{x}-2\right)+\frac{1}{4}{(\mathrm{x}-2)}^2-\frac{1}{12}{(\mathrm{x}-2)}^3+...\right)$

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.

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

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

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

Let $\mathrm{x}-2=\mathrm{t}$:

$\begin{array}{l}(\mathrm{t}+2{)}^2\sum _{\mathrm{n}=2}^{\infty }\mathrm{n}(\mathrm{n}-1){\mathrm{a}}_{\mathrm{n}}(\mathrm{t}{)}^{\mathrm{n}-2}-\mathrm{x}\sum _{\mathrm{n}=2}^{\infty }{\mathrm{na}}_{\mathrm{n}}(\mathrm{x}-2{)}^{\mathrm{n}-1}+2\sum _{\mathrm{n}=0}^{\infty }{\mathrm{a}}_{\mathrm{n}}{\left(\mathrm{x}-2\right)}^{\mathrm{n}}=0\\ \left(2{\mathrm{a}}_2{\mathrm{t}}^2+6{\mathrm{a}}_3{\mathrm{t}}^3+12{\mathrm{a}}_4{\mathrm{t}}^4+20{\mathrm{a}}_5{\mathrm{t}}^5+...\right)+\left(8{\mathrm{a}}_2\mathrm{t}+24{\mathrm{a}}_3{\mathrm{t}}^3+48{\mathrm{a}}_4{\mathrm{t}}^3+80{\mathrm{a}}_5{\mathrm{t}}^4+...\right)+(8{\mathrm{a}}_2+24{\mathrm{a}}_3{\mathrm{t}}^3+48{\mathrm{a}}_4{\mathrm{t}}^2+80{\mathrm{a}}_5{\mathrm{t}}^3+...)+\\ (-{\mathrm{a}}_1\mathrm{t}-2{\mathrm{a}}_2{\mathrm{t}}^2-3{\mathrm{a}}_3{\mathrm{t}}^3-4{\mathrm{a}}_4{\mathrm{t}}^4+...)+(-2{\mathrm{a}}_1-4{\mathrm{a}}_2\mathrm{t}-6{\mathrm{a}}_3{\mathrm{t}}^2-4{\mathrm{a}}_4{\mathrm{t}}^4+...)+\\ (-2{\mathrm{a}}_1-4{\mathrm{a}}_2\mathrm{t}-6{\mathrm{a}}_3{\mathrm{t}}^2-8{\mathrm{a}}_4{\mathrm{t}}^3+...)+(2{\mathrm{a}}_0+2{\mathrm{a}}_1\mathrm{t}+2{\mathrm{a}}_2{\mathrm{t}}^2+2{\mathrm{a}}_3{\mathrm{t}}^3+2{\mathrm{a}}_4{\mathrm{t}}^4+...)=0\\ \left(2{\mathrm{a}}_0-2{\mathrm{a}}_1+8{\mathrm{a}}_2\right)+\left(8{\mathrm{a}}_2+24{\mathrm{a}}_3-{\mathrm{a}}_1-4{\mathrm{a}}_2+2{\mathrm{a}}_1\right)\mathrm{t}+...=0\\ 2{\mathrm{a}}_0-2{\mathrm{a}}_1+8{\mathrm{a}}_2=0\Rightarrow {\mathrm{a}}_2=\frac{{\mathrm{a}}_1-{\mathrm{a}}_0}{4}\\ 8{\mathrm{a}}_2+24{\mathrm{a}}_3-{\mathrm{a}}_1-4{\mathrm{a}}_2+2{\mathrm{a}}_1=0\Rightarrow {\mathrm{a}}_3=\frac{{\mathrm{a}}_0-2{\mathrm{a}}_1}{24}\end{array}$

At the end,

$\begin{array}{l}\mathrm{y}=\sum _{\mathrm{n}=0}^{\infty }{\mathrm{a}}_{\mathrm{n}}{\left(\mathrm{x}-2\right)}^{\mathrm{n}}={\mathrm{a}}_0+{\mathrm{a}}_1(\mathrm{x}-2)+{\mathrm{a}}_2(\mathrm{x}-2{)}^2+{\mathrm{a}}_3(\mathrm{x}-2{)}_3+...\\ \mathrm{y}={\mathrm{a}}_0\left(1+\frac{1}{4}{(\mathrm{x}-2)}^2+\frac{1}{24}{(\mathrm{x}-2)}^3+...\right)+{\mathrm{a}}_1\left(\left(\mathrm{x}-2\right)+\frac{1}{4}{(\mathrm{x}-2)}^2-\frac{1}{12}{(\mathrm{x}-2)}^3+...\right)\end{array}$

Hence, the final answer is:

$\mathrm{y}={\mathrm{a}}_0\left(1+\frac{1}{4}{(\mathrm{x}-2)}^2+\frac{1}{24}{(\mathrm{x}-2)}^3+...\right)+{\mathrm{a}}_1\left(\left(\mathrm{x}-2\right)+\frac{1}{4}{(\mathrm{x}-2)}^2-\frac{1}{12}{(\mathrm{x}-2)}^3+...\right)$