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,

$$\begin{gathered} \mathrm{y}\text{'}\text{'}-\left(\mathrm{cosx}\right)\mathrm{y}\text{'}-\mathrm{y}=0 \\ \mathrm{y}(\mathrm{Ï€}/2)=1,\mathrm{y}\text{'}(\mathrm{Ï€}/2)=1 \end{gathered}$$

Apply a substitution and transform the equation,

$$\begin{gathered} \mathrm{y}\text{'}\text{'}-\cos \left(\mathrm{t}+\frac{\mathrm{π}}{2}\right)·\mathrm{y}\text{'}-\mathrm{y}=0 \\ \mathrm{y}\text{'}\text{'}-\mathrm{sint}·\mathrm{y}\text{'}-\mathrm{y}=0 \end{gathered}$$

Use the formula

$$\begin{gathered} \mathrm{Y}\left(\mathrm{x}\right)=\underset{\mathrm{n}=0}{\overset{∞}{∑}}{\mathrm{a}}_{\mathrm{n}}{\mathrm{t}}^{\mathrm{n}} \\ \mathrm{Y}\text{'}\left(\mathrm{t}\right)=\underset{\mathrm{n}=1}{\overset{∞}{∑}}\mathrm{n}·{\mathrm{a}}_{\mathrm{n}}{\left(\mathrm{t}\right)}^{\mathrm{n}-1} \\ \mathrm{Y}\text{'}\text{'}\left(\mathrm{t}\right)=\underset{\mathrm{n}=2}{\overset{∞}{∑}}\mathrm{n}(\mathrm{n}-1)·{\mathrm{a}}_{\mathrm{n}}{\left(\mathrm{t}\right)}^{\mathrm{n}-2} \end{gathered}$$

Substitute it in the above equation we get,

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

Hence we get the relation:

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

The series expansion for the function is

$$\begin{gathered} \left(2{\mathrm{a}}_2+6{\mathrm{a}}_3\mathrm{t}+12{\mathrm{a}}_4{\mathrm{t}}^2+20{\mathrm{a}}_5{\mathrm{t}}^3+⋯\right)-\left({\mathrm{a}}_1\mathrm{t}+2{\mathrm{a}}_2{\mathrm{t}}^2+3{\mathrm{a}}_3{\mathrm{t}}^3+4{\mathrm{a}}_4{\mathrm{t}}^4+⋯\right)+\left({\mathrm{a}}_1\frac{{\mathrm{t}}^3}{3!}+2{\mathrm{a}}_2\frac{{\mathrm{t}}^4}{3!}+3{\mathrm{a}}_3\frac{{\mathrm{t}}^5}{3!}+4{\mathrm{a}}_4\frac{{\mathrm{t}}^6}{3!}+⋯\right) \\ -\left({\mathrm{a}}_1\frac{{\mathrm{t}}^5}{5!}+2{\mathrm{a}}_2\frac{{\mathrm{t}}^6}{5!}+3{\mathrm{a}}_3\frac{{\mathrm{t}}^7}{5!}+4{\mathrm{a}}_4\frac{{\mathrm{t}}^8}{5!}+⋯\right)+⋯-\left({\mathrm{a}}_0+{\mathrm{a}}_1\mathrm{t}+{\mathrm{a}}_2{\mathrm{t}}^2+{\mathrm{a}}_3{\mathrm{t}}^3+{\mathrm{a}}_4{\mathrm{t}}^4+⋯\right)=0 \end{gathered}$$

Taking coefficients and exponents of the same power.

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

Simplify the expression:

$\left(2{\mathrm{a}}_2-{\mathrm{a}}_0\right)+\left(6{\mathrm{a}}_3-{\mathrm{a}}_1-{\mathrm{a}}_1\right)\left(\mathrm{x}-\frac{\mathrm{π}}{2}\right)+\left(12{\mathrm{a}}_4-2{\mathrm{a}}_2-{\mathrm{a}}_2\right){\left(\mathrm{x}-\frac{\mathrm{π}}{2}\right)}^2+\left(20{\mathrm{a}}_5-3{\mathrm{a}}_3+\frac{{\mathrm{a}}_1}{6}-{\mathrm{a}}_3\right){\left(\mathrm{x}-\frac{\mathrm{π}}{2}\right)}^3+⋯=0$

Hence, the expression after the expansion is:

$\left(2{\mathrm{a}}_2-{\mathrm{a}}_0\right)+\left(6{\mathrm{a}}_3-{\mathrm{a}}_1-{\mathrm{a}}_1\right)\left(\mathrm{x}-\frac{\mathrm{π}}{2}\right)+\left(12{\mathrm{a}}_4-2{\mathrm{a}}_2-{\mathrm{a}}_2\right){\left(\mathrm{x}-\frac{\mathrm{π}}{2}\right)}^2+\left(20{\mathrm{a}}_5-3{\mathrm{a}}_3+\frac{{\mathrm{a}}_1}{6}-{\mathrm{a}}_3\right){\left(\mathrm{x}-\frac{\mathrm{π}}{2}\right)}^3+⋯=0$

By equating the coefficients, we get,

$$\begin{gathered} 2{\mathrm{a}}_2-{\mathrm{a}}_0=0→{\mathrm{a}}_2=\frac{{\mathrm{a}}_0}{2}=\frac{1}{2} \\ 6{\mathrm{a}}_3-{\mathrm{a}}_1-{\mathrm{a}}_1→{\mathrm{a}}_3=\frac{{\mathrm{a}}_1}{3}=\frac{1}{3} \end{gathered}$$

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.

$\mathrm{Y}\left(\mathrm{x}\right)=1+\left(\mathrm{x}-\frac{\mathrm{π}}{2}\right)+\frac{1}{2}{\left(\mathrm{x}-\frac{\mathrm{π}}{2}\right)}^2+\frac{1}{3}{\left(\mathrm{x}-\frac{\mathrm{π}}{2}\right)}^3+⋯$

Hence, the first four nonzero terms are:

$\mathrm{Y}\left(\mathrm{x}\right)=1+\left(\mathrm{x}-\frac{\mathrm{π}}{2}\right)+\frac{1}{2}{\left(\mathrm{x}-\frac{\mathrm{π}}{2}\right)}^2+\frac{1}{3}{\left(\mathrm{x}-\frac{\mathrm{π}}{2}\right)}^3+⋯$