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,

$$\begin{gathered} \mathrm{y}\text{'}\text{'}-{\mathrm{e}}^{2\mathrm{x}}\mathrm{y}\text{'}+\left(\mathrm{cosx}\right)\mathrm{y}=0 \\ \mathrm{y}\left(0\right)=-1,\mathrm{y}\text{'}\left(0\right)=1 \end{gathered}$$

Use the formula

$$\begin{gathered} \mathrm{Y}\left(\mathrm{x}\right)=\underset{\mathrm{n}=0}{\overset{∞}{∑}}{\mathrm{a}}_{\mathrm{n}}{\mathrm{t}}^{\mathrm{n}} \\ \mathrm{Y}\text{'}\left(\mathrm{t}\right)=\underset{\mathrm{n}=1}{\overset{∞}{∑}}\mathrm{n}·{\mathrm{a}}_{\mathrm{n}}{\left(\mathrm{x}\right)}^{\mathrm{n}-1} \\ \mathrm{Y}\text{'}\text{'}\left(\mathrm{t}\right)=\underset{\mathrm{n}=2}{\overset{∞}{∑}}\mathrm{n}(\mathrm{n}-1)·{\mathrm{a}}_{\mathrm{n}}{\left(\mathrm{x}\right)}^{\mathrm{n}-2} \\ \mathrm{cosx}=1-\frac{{\mathrm{x}}^2}{2}+\frac{{\mathrm{x}}^4}{24}-⋯ \\ {\mathrm{e}}^{2\mathrm{x}}=1+2\mathrm{x}+2{\mathrm{x}}^2+\frac{4}{3}{\mathrm{x}}^3+⋯ \end{gathered}$$

Substitute it in the above equation we get,

role="math" localid="1664100676580" $\underset{\mathrm{n}=2}{\overset{∞}{∑}}\mathrm{n}(\mathrm{n}-1)·{\mathrm{a}}_{\mathrm{n}}{\left(\mathrm{x}\right)}^{\mathrm{n}-2}-\left(1+2\mathrm{x}+2{\mathrm{x}}^2+\frac{4}{3}{\mathrm{x}}^3+⋯\right)\underset{\mathrm{n}=1}{\overset{∞}{∑}}\mathrm{n}·{\mathrm{a}}_{\mathrm{n}}{\left(\mathrm{x}\right)}^{\mathrm{n}-1}-\left(1-\frac{{\mathrm{x}}^2}{2}+\frac{{\mathrm{x}}^4}{24}+⋯\right)\underset{\mathrm{n}=0}{\overset{∞}{∑}}{\mathrm{a}}_{\mathrm{n}}{\left(\mathrm{t}\right)}^{\mathrm{n}}=0$

Hence, we get the relation:$\underset{\mathrm{n}=2}{\overset{∞}{∑}}\mathrm{n}(\mathrm{n}-1)·{\mathrm{a}}_{\mathrm{n}}{\left(\mathrm{x}\right)}^{\mathrm{n}-2}-\left(1+2\mathrm{x}+2{\mathrm{x}}^2+\frac{4}{3}{\mathrm{x}}^3+⋯\right)\underset{\mathrm{n}=1}{\overset{∞}{∑}}\mathrm{n}·{\mathrm{a}}_{\mathrm{n}}{\left(\mathrm{x}\right)}^{\mathrm{n}-1}-\left(1-\frac{{\mathrm{x}}^2}{2}+\frac{{\mathrm{x}}^4}{24}+⋯\right)\underset{\mathrm{n}=0}{\overset{∞}{∑}}{\mathrm{a}}_{\mathrm{n}}{\left(\mathrm{t}\right)}^{\mathrm{n}}=0$.

The series expansion for the function is:

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

Taking coefficients and exponents of the same power. Simplify the expression:

$\left(2{\mathrm{a}}_2-{\mathrm{a}}_1+{\mathrm{a}}_0\right)+\left(6{\mathrm{a}}_3-2{\mathrm{a}}_2-2{\mathrm{a}}_1+{\mathrm{a}}_1\right)\mathrm{x}+\left(12{\mathrm{a}}_4-3{\mathrm{a}}_3+4{\mathrm{a}}_2-2{\mathrm{a}}_1+{\mathrm{a}}_2-\frac{{\mathrm{a}}_0}{2}\right){\mathrm{x}}^2+⋯=0$

Hence the expression after the expansion is:

$\left(2{\mathrm{a}}_2-{\mathrm{a}}_1+{\mathrm{a}}_0\right)+\left(6{\mathrm{a}}_3-2{\mathrm{a}}_2-2{\mathrm{a}}_1+{\mathrm{a}}_1\right)\mathrm{x}+\left(12{\mathrm{a}}_4-3{\mathrm{a}}_3+4{\mathrm{a}}_2-2{\mathrm{a}}_1+{\mathrm{a}}_2-\frac{{\mathrm{a}}_0}{2}\right){\mathrm{x}}^2+⋯=0$

By equating the coefficients, we get,

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

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.

role="math" localid="1664101022601" $\mathrm{y}\left(\mathrm{x}\right)=-1+\mathrm{x}+{\mathrm{x}}^2+\frac{{\mathrm{x}}^3}{2}+⋯$

Hence, the first four nonzero terms are $\mathrm{y}\left(\mathrm{x}\right)=-1+\mathrm{x}+{\mathrm{x}}^2+\frac{{\mathrm{x}}^3}{2}+⋯$.