91Ó°ÊÓ

The power series approach is used in mathematics to find a power series solution to certain differential equations. In general, such a solution starts with an unknown power series and then plugs that solution into the differential equation to obtain a coefficient recurrence relation.

A differential equation's power series solution is a function with an infinite number of terms, each holding a different power of the dependent variable. It is generally given by the formula,

$\mathbf{y}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\underset{\mathbf{n}\mathbf{=}\mathbf{0}}{\overset{\mathbf{∞}}{\mathbf{∑}}}{\mathbf{a}}_{\mathbf{n}}{\mathbf{x}}^{\mathbf{n}}$

Given,

$\mathrm{y}\text{'}\text{'}-\left(\mathrm{sinx}\right)\mathrm{y}=0;\mathrm{y}\left(\mathrm{Ï€}\right)=1,\mathrm{y}\text{'}\left(\mathrm{Ï€}\right)=0$

Apply a substitution and transform the equation,

$$\begin{gathered} \mathrm{x}=\mathrm{t}+\mathrm{Ï€} \\ \mathrm{y}\left(\mathrm{x}\right)=\mathrm{y}(\mathrm{t}+\mathrm{Ï€})=\mathrm{Y}\left(\mathrm{t}\right) \\ \mathrm{Y}\text{'}\text{'}-\sin (\mathrm{t}+\mathrm{Ï€})\mathrm{Y}=0,\mathrm{Y}\left(0\right)=1,\mathrm{Y}\text{'}\left(0\right)=0 \end{gathered}$$

We know,

$\sin (\mathrm{x}+\mathrm{Ï€})=-\mathrm{sinx}$

We get,

$$\begin{gathered} -\sin (\mathrm{t}+\mathrm{Ï€})=\mathrm{²õ¾\pm ²ÔÏ€} \\ \mathrm{Y}\text{'}\text{'}+\sin \left(\mathrm{t}\right)\mathrm{Y}=0,\mathrm{Y}\left(0\right)=1,\mathrm{Y}\text{'}\left(0\right)=0 \\ \mathrm{p}\left(\mathrm{t}\right)=\mathrm{sint},\mathrm{q}\left(\mathrm{t}\right)=0 \end{gathered}$$

Use the formula

role="math" localid="1664097188824" $$\begin{gathered} \mathrm{Y}\left(\mathrm{x}\right)=\underset{\mathrm{n}=0}{\overset{∞}{∑}}{\mathrm{a}}_{\mathrm{n}}{\mathrm{t}}^{\mathrm{n}} \\ \mathrm{Y}\text{'}\text{'}\left(\mathrm{x}\right)=\underset{\mathrm{n}=2}{\overset{∞}{∑}}\mathrm{n}(\mathrm{n}-1){\mathrm{a}}_{\mathrm{n}}{\mathrm{t}}^{\mathrm{n}-2} \\ \underset{\mathrm{n}=2}{\overset{∞}{∑}}\mathrm{n}(\mathrm{n}-1){\mathrm{a}}_{\mathrm{n}}{\mathrm{t}}^{\mathrm{n}-2}+\mathrm{sint}\underset{\mathrm{n}=2}{\overset{∞}{∑}}{\mathrm{a}}_{\mathrm{n}}{\mathrm{t}}^{\mathrm{n}}=0 \end{gathered}$$

$\mathrm{sint}=\mathrm{t}-\frac{{\mathrm{t}}^3}{3!}+\frac{{\mathrm{t}}^5}{5!}-\frac{{\mathrm{t}}^7}{7!}+…$

Substitute it in the above equation we get,

$\underset{\mathrm{n}=2}{\overset{∞}{∑}}\mathrm{n}(\mathrm{n}-1){\mathrm{a}}_{\mathrm{n}}{\mathrm{t}}^{\mathrm{n}-2}+\left(\mathrm{t}-\frac{{\mathrm{t}}^3}{3!}+\frac{{\mathrm{t}}^5}{5!}-\frac{{\mathrm{t}}^7}{7!}+…\right)\underset{\mathrm{n}=0}{\overset{∞}{∑}}{\mathrm{a}}_{\mathrm{n}}{\mathrm{t}}^{\mathrm{n}}=0$

Hence we get the relation $\underset{\mathrm{n}=2}{\overset{∞}{∑}}\mathrm{n}(\mathrm{n}-1){\mathrm{a}}_{\mathrm{n}}{\mathrm{t}}^{\mathrm{n}-2}+\left(\mathrm{t}-\frac{{\mathrm{t}}^3}{3!}+\frac{{\mathrm{t}}^5}{5!}-\frac{{\mathrm{t}}^7}{7!}+…\right)\underset{\mathrm{n}=0}{\overset{∞}{∑}}{\mathrm{a}}_{\mathrm{n}}{\mathrm{t}}^{\mathrm{n}}=0$.

The series expansion for the function is

$$\begin{gathered} \underset{\mathrm{n}=2}{\overset{∞}{∑}}\mathrm{n}(\mathrm{n}-1){\mathrm{a}}_{\mathrm{n}}{\mathrm{t}}^{\mathrm{n}-2}=2(2-1){\mathrm{a}}_2{\mathrm{t}}^{2-2}+3(3-1){\mathrm{a}}_3{\mathrm{t}}^{3-2}+4(4-1){\mathrm{a}}_4{\mathrm{t}}^{4-2}+5(5-1){\mathrm{a}}_5{\mathrm{t}}^{5-2}+… \\ \underset{\mathrm{n}=2}{\overset{∞}{∑}}\mathrm{n}(\mathrm{n}-1){\mathrm{a}}_{\mathrm{n}}{\mathrm{t}}^{\mathrm{n}-2}=2{\mathrm{a}}_2+6{\mathrm{a}}_3{\mathrm{t}}^1+12{\mathrm{a}}_4{\mathrm{t}}^2+20{\mathrm{a}}_5{\mathrm{t}}^3… \\ \underset{\mathrm{n}=2}{\overset{∞}{∑}}{\mathrm{a}}_{\mathrm{n}}{\mathrm{t}}^{\mathrm{n}}={\mathrm{a}}_0+{\mathrm{a}}_1\mathrm{t}+{\mathrm{a}}_2{\mathrm{t}}^2+{\mathrm{a}}_3{\mathrm{t}}^3+… \end{gathered}$$

By expanding the series we get,

$$\begin{gathered} \left(\mathrm{t}-\frac{{\mathrm{t}}^3}{3!}+\frac{{\mathrm{t}}^5}{5!}-\frac{{\mathrm{t}}^7}{7!}+…\right)\underset{\mathrm{n}=0}{\overset{∞}{∑}}{\mathrm{a}}_{\mathrm{n}}{\mathrm{t}}^{\mathrm{n}+1}=\left(\mathrm{t}-\frac{{\mathrm{t}}^3}{3!}+\frac{{\mathrm{t}}^5}{5!}-\frac{{\mathrm{t}}^7}{7!}+…\right)\left({\mathrm{a}}_0+{\mathrm{a}}_1\mathrm{t}+{\mathrm{a}}_2{\mathrm{t}}^2+{\mathrm{a}}_3{\mathrm{t}}^3+…\right) \\ \left(\mathrm{t}-\frac{{\mathrm{t}}^3}{3!}+\frac{{\mathrm{t}}^5}{5!}-\frac{{\mathrm{t}}^7}{7!}+…\right)\underset{\mathrm{n}=0}{\overset{∞}{∑}}{\mathrm{a}}_{\mathrm{n}}{\mathrm{t}}^{\mathrm{n}+1}=\mathrm{t}·{\mathrm{a}}_0+{\mathrm{t}}^2·{\mathrm{a}}_1+{\mathrm{t}}^3·{\mathrm{a}}_2-\frac{{\mathrm{t}}^3}{3!}·{\mathrm{a}}_0… \end{gathered}$$

We need only the first four terms. Sum up the equation.

$2{\mathrm{a}}_2+6{\mathrm{a}}_3{\mathrm{t}}^1+12{\mathrm{a}}_4{\mathrm{t}}^2+20{\mathrm{a}}_5{\mathrm{t}}^3+…+\mathrm{t}·{\mathrm{a}}_0+{\mathrm{t}}^2·{\mathrm{a}}_1+{\mathrm{t}}^3·{\mathrm{a}}_2…=0$

Simplify the expression.

$2{\mathrm{a}}_2+\left(6{\mathrm{a}}_3+{\mathrm{a}}_0\right){\mathrm{t}}^1+\left(12{\mathrm{a}}_4+{\mathrm{a}}_1\right){\mathrm{t}}^2+\left({\mathrm{a}}_2+20{\mathrm{a}}_5-\frac{{\mathrm{a}}_0}{3!}\right){\mathrm{t}}^3+…=0$

Hence the expression after the expansion is:

$2{\mathrm{a}}_2+\left(6{\mathrm{a}}_3+{\mathrm{a}}_0\right){\mathrm{t}}^1+\left(12{\mathrm{a}}_4+{\mathrm{a}}_1\right){\mathrm{t}}^2+\left({\mathrm{a}}_2+20{\mathrm{a}}_5-\frac{{\mathrm{a}}_0}{3!}\right){\mathrm{t}}^3+…=0$

By equating the coefficients, we get,

$$\begin{gathered} 2{\mathrm{a}}_2=0→{\mathrm{a}}_2=0 \\ 6{\mathrm{a}}_3+{\mathrm{a}}_0=0→{\mathrm{a}}_3=-\frac{{\mathrm{a}}_0}{6} \\ 12{\mathrm{a}}_4+{\mathrm{a}}_1=0→{\mathrm{a}}_4=-\frac{{\mathrm{a}}_1}{12} \\ {\mathrm{a}}_2+20{\mathrm{a}}_5-\frac{{\mathrm{a}}_0}{6}=0→{\mathrm{a}}_5=\frac{{\mathrm{a}}_0}{120} \end{gathered}$$

The general solution was:

$\mathrm{y}\left(\mathrm{t}\right)=\underset{\mathrm{n}=0}{\overset{∞}{∑}}{\mathrm{a}}_{\mathrm{n}}{\mathrm{t}}^{\mathrm{n}}={\mathrm{a}}_0+{\mathrm{a}}_1\mathrm{t}+{\mathrm{a}}_2{\mathrm{t}}^2+{\mathrm{a}}_3{\mathrm{t}}^3+⋯$

Apply the initial condition and substitute the coefficient.

$$\begin{gathered} \mathrm{Y}\left(\mathrm{t}\right)={\mathrm{a}}_0+{\mathrm{a}}_1\mathrm{t}+{\mathrm{a}}_2{\mathrm{t}}^2+{\mathrm{a}}_3{\mathrm{t}}^3+{\mathrm{a}}_4{\mathrm{t}}^4+{\mathrm{a}}_5{\mathrm{t}}^5… \\ \mathrm{Y}\left(\mathrm{t}\right)={\mathrm{a}}_0+{\mathrm{a}}_1\mathrm{t}-\frac{1}{6}{\mathrm{a}}_0{\mathrm{t}}^3-\frac{1}{12}{\mathrm{a}}_1{\mathrm{t}}^4+\frac{1}{120}{\mathrm{a}}_0{\mathrm{t}}^5+… \\ \mathrm{Y}\left(\mathrm{t}\right)=1-\frac{1}{6}{\mathrm{t}}^3+\frac{1}{120}{\mathrm{t}}^5+… \\ \mathrm{Y}\left(\mathrm{x}\right)=1-\frac{1}{6}(\mathrm{x}-\mathrm{π}{)}^3+\frac{1}{120}(\mathrm{x}-\mathrm{π}{)}^5+… \end{gathered}$$

Hence, the first four nonzero terms are$\mathrm{Y}\left(\mathrm{x}\right)=1-\frac{1}{6}(\mathrm{x}-\mathrm{π}{)}^3+\frac{1}{120}(\mathrm{x}-\mathrm{π}{)}^5+…$.