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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,

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

Given,

$\mathrm{y}\text{'}-{\mathrm{e}}^{\mathrm{x}}\mathrm{y}=0;\mathrm{y}\left(0\right)=1$

From the above equation put$\mathrm{p}\left(\mathrm{x}\right)={\mathrm{e}}^{\mathrm{x}}$ which is analytic over the entire number line.

Use the formula,

$\mathrm{y}\left(\mathrm{x}\right)=\underset{\mathrm{n}=0}{\overset{鈭�}{鈭�}}{\mathrm{a}}_{\mathrm{n}}{\mathrm{x}}^{\mathrm{n}}$

Taking derivative and substituting in the equation, we get the relation,

$$\begin{gathered} \mathrm{y}\text{'}\left(\mathrm{x}\right)=\underset{\mathrm{n}=1}{\overset{鈭�}{鈭�}}\mathrm{n}路{\mathrm{a}}_{\mathrm{n}}{\left(\mathrm{x}\right)}^{\mathrm{n}-1} \\ \underset{\mathrm{n}=1}{\overset{鈭�}{鈭�}}\mathrm{n}路{\mathrm{a}}_{\mathrm{n}}{\left(\mathrm{x}\right)}^{\mathrm{n}-1}-{\mathrm{e}}^{\mathrm{x}}\underset{\mathrm{n}=1}{\overset{鈭�}{鈭�}}\mathrm{n}路{\mathrm{a}}_{\mathrm{n}}{\left(\mathrm{x}\right)}^{\mathrm{n}}=0 \end{gathered}$$

The series expansion for the function is

${\mathrm{e}}^{\mathrm{x}}=1+\mathrm{x}+\frac{{\mathrm{x}}^2}{2!}+\frac{{\mathrm{x}}^3}{3!}+\frac{{\mathrm{x}}^4}{4!}+鈰�$

By expanding the series we get,

$\left({\mathrm{a}}_1+2{\mathrm{a}}_2\mathrm{x}+3{\mathrm{a}}_3{\mathrm{x}}^2+4{\mathrm{a}}_4{\mathrm{x}}^3鈰�\right)-\left(1+\mathrm{x}+\frac{{\mathrm{x}}^2}{2!}+\frac{{\mathrm{x}}^3}{3!}+\frac{{\mathrm{x}}^4}{4!}+鈰�\right)\left({\mathrm{a}}_0+{\mathrm{a}}_1\mathrm{x}+{\mathrm{a}}_2{\mathrm{x}}^2+{\mathrm{a}}_3{\mathrm{x}}^3+鈰�\right)=0$

Hence the expression after the expansion is:

$\left({\mathrm{a}}_1+2{\mathrm{a}}_2\mathrm{x}+3{\mathrm{a}}_3{\mathrm{x}}^2+4{\mathrm{a}}_4{\mathrm{x}}^3鈰�\right)-\left(1+\mathrm{x}+\frac{{\mathrm{x}}^2}{2!}+\frac{{\mathrm{x}}^3}{3!}+\frac{{\mathrm{x}}^4}{4!}+鈰�\right)\left({\mathrm{a}}_0+{\mathrm{a}}_1\mathrm{x}+{\mathrm{a}}_2{\mathrm{x}}^2+{\mathrm{a}}_3{\mathrm{x}}^3+鈰�\right)=0$

Expand the expression given in the previous step.

$$\begin{gathered} \left(a_1+2a_{2}x+3a_3{x}^2+\right.\left.4a_4{x}^3+5a_5{x}^4+鈰�\right)+\left(-a_0-a_{1}x-a_2{x}^2-a_3{x}^3-a_4{x}^4+鈰�\right) \\ +\left(-a_{0}x-a_1{x}^2-a_2{x}^3-a_3{x}^4-a_4{x}^5+鈰�\right)+\left(-a_0\frac{x^2}{2}-a_1\frac{x^3}{2}-a_2\frac{x^4}{2}-a_3\frac{x^5}{2}-鈰�\right) \end{gathered}$$

$\left({\mathrm{a}}_1-{\mathrm{a}}_0\right)+\left(2{\mathrm{a}}_2-{\mathrm{a}}_1-{\mathrm{a}}_0\right)\mathrm{x}+\left(3{\mathrm{a}}_3-{\mathrm{a}}_2-{\mathrm{a}}_1-\frac{{\mathrm{a}}_0}{2}\right){\mathrm{x}}^2+鈰�=0$

By equating the coefficients,

$$\begin{gathered} {\mathrm{a}}_1-{\mathrm{a}}_0=0鈫�{\mathrm{a}}_1={\mathrm{a}}_0 \\ 2{\mathrm{a}}_2-{\mathrm{a}}_1-{\mathrm{a}}_0=0鈫�{\mathrm{a}}_2={\mathrm{a}}_0 \\ 3{\mathrm{a}}_3-{\mathrm{a}}_2-{\mathrm{a}}_1-\frac{{\mathrm{a}}_0}{2}=0鈫�{\mathrm{a}}_3=\frac{5}{6}{\mathrm{a}}_0 \end{gathered}$$

The general solution was

$\mathrm{y}\left(\mathrm{x}\right)=\underset{\mathrm{n}=0}{\overset{鈭�}{鈭�}}{\mathrm{a}}_{\mathrm{n}}{\left(\mathrm{x}\right)}^{\mathrm{n}}={\mathrm{a}}_0+{\mathrm{a}}_1\left(\mathrm{x}\right)+{\mathrm{a}}_2(\mathrm{x}{)}^2+{\mathrm{a}}_3(\mathrm{x}{)}^3+鈰�$

Apply the initial condition and substitute the coefficient.

$\mathrm{y}\left(\mathrm{x}\right)=1+\mathrm{x}+{\mathrm{x}}^2+\frac{5}{6}{\mathrm{x}}^3+鈰�$

Hence, the first four nonzero terms are$\mathrm{y}\left(\mathrm{x}\right)=1+\mathrm{x}+{\mathrm{x}}^2+\frac{5}{6}{\mathrm{x}}^3+鈰�$.