Question

In Problems 23 and 24 use the procedure in Example 8 to find two power series solutions of the given differential equation about the ordinary point

$\begin{array}{rcl}x& =& 0\\ y^{\text{'}\text{'}}+e^{\text{'}}{y}^{\text{'}}-y& =& 0\end{array}$

Short answer

Therefore, the two linearly independent solutions are;

$$\begin{gathered} y_1=1+\frac{1}{2}{x}^2-\frac{1}{6}{x}^3+\dots \\ {y}_2=x-\frac{1}{2}{x}^2+\frac{1}{6}{x}^3-\frac{1}{24}{x}^4+\dots \end{gathered}$$

Step-by-step solution

Given Information

The given differential equation:

$y^{\text{'}\text{'}}+e^{\text{'}}{y}^{\text{'}}-y=0$

Plug into the differential equation

$f\left(y^{\text{'}\text{'}},y^{\text{'}},y\right)=y^{\text{'}\text{'}}+e^x{y}^{\text{'}}-y=0$

Which has an ordinary point at $x=0$We state that, without proof, if the differential equation has an ordinary point at$x=x_0,$then it has a power series solution, with two linearly independent solutions, which has the following form:

$y=\sum _{n=0}^{\infty }{c}_n{\left(x-x_0\right)}^n$

Thus, the given differential equation has a solution of the following form,

$y=\sum _{n=0}^{\infty }{c}_n{x}^n \dots \left(1\right),$Where $x_0=0$

We need to find the two linearly independent series solutions for the given differential equation. We find the first and second derivatives of (1) as follows:

$$\begin{gathered} y\text{'}=\sum _{n=0}^{\infty }{c}_{n}n{x}^{n-1} \\ y\text{'}\text{'}=\sum _{n=0}^{\infty }{c}_{n}n(n-1)x^{n-2} \end{gathered}$$

We shift the summation index to $n=1$for $y\text{'}.$like the first term of the power series equals to zero, and $n=2$for$y".$

We use,

$$\begin{gathered} y\text{'}=\sum _{n=1}^{\infty }{c}_{n}n{x}^{n-1} \dots \left(2\right) \\ y\text{'}\text{'}=\sum _{n=2}^{\infty }{c}_{n}n(n-1)x^{n-2} \dots \left(3\right) \end{gathered}$$

Substituting into the given differential equation

By substituting (1), (2), and (3) into the given differential equation, yields

$\begin{array}{rcl}f\left(y^{\text{'}\text{'}},y^{\text{'}},y\right)& =& \sum _{n=2}^{\infty }{c}_{n}n(n-1)x^{n-2}+e^x\sum _{n=1}^{\infty }{c}_{n}n{x}^{n-1}-\sum _{n=0}^{\infty }{c}_n{x}^n\\ & =& \sum _{n=2}^{\infty }{c}_{n}n(n-1)x^{n-2}+\sum _{n=0}^{\infty }\frac{1}{n!}{x}^n\sum _{n=1}^{\infty }{c}_{n}n{x}^{n-1}-\sum _{n=0}^{\infty }{c}_n{x}^n\\ & & \end{array}$

We expand the four series above and yields

$\begin{array}{rcl}f\left(y^{\text{'}\text{'}},y^{\text{'}},y\right)& =& \left(2c_2+6c_{3}x+12c_4{x}^2+20c_5{x}^3+\dots \right)+\left(1+x+\frac{1}{2}{x}^2+\frac{1}{6}{x}^3+\dots \right)\\ & & \left(c_1+2c_{2}x+3c_3{x}^2\right.\left.+4c_4{x}^3+5c_5{x}^4+\dots \right)-\left(c_0+c_{1}x+c_2{x}^2+c_3{x}^3+\dots \right)\\ & =& \left(2c_2+6c_{3}x+12c_4{x}^2+20c_5{x}^3+\dots \right)+1·\left(c_1+2c_{2}x+3c_3{x}^2+4c_4{x}^3+5c_5{x}^4+\dots \right)\\ & & +x·\left(c_1+2c_{2}x+3c_3{x}^2+4c_4{x}^3+\dots \right)+\frac{1}{2}{x}^2·\left(c_1+2c_{2}x+3c_3{x}^2+4c_4{x}^3+\dots \right)\\ & & +\frac{1}{6}{x}^3·\left(c_1+2c_{2}x+3c_3{x}^2+4c_4{x}^3+\dots \right)-\left(c_0+c_{1}x+c_2{x}^2+c_3{x}^3+\dots \right)+\dots \\ & =& \left(2c_2+6c_{3}x+12c_4{x}^2+20c_5{x}^3+\dots \right)+\left(c_1+2c_{2}x+3c_3{x}^2+4c_4{x}^3+5c_5{x}^4+\dots \right)\\ & & +\left(c_{1}x+2c_2{x}^2+3c_3{x}^3+4c_4{x}^4+\dots \right)+\left(\frac{1}{2}{c}_1{x}^2+c_2{x}^3+\frac{3}{2}{c}_3{x}^4+\frac{1}{2}{c}_4{x}^5+\dots \right)\\ & & +\left(\frac{1}{6}{c}_1{x}^3+\frac{1}{3}{c}_2{x}^4+\frac{1}{2}{c}_3{x}^5+\frac{2}{3}{c}_4{x}^6+\dots \right)-\left(c_0+c_{1}x+c_2{x}^2+c_3{x}^3+\dots \right)+\dots \\ & & \end{array}$

Find the coefficients of the terms

Now, we sum the coefficients of the terms, which are raised to the same power together, yields:

$\begin{array}{rcl}f\left(y^{\text{'}\text{'}},y^{\text{'}},y\right)& =& \left(2c_2+c_1-c_0\right)+\left(6c_3+2c_2+c_1-c_1\right)x+\left(12c_4+3c_3+2c_2+\frac{1}{2}{c}_1-c_2\right)x^2\\ & & +\left(20c_5+4c_4+3c_3+c_2+\frac{1}{6}{c}_1-c_3\right)x^3+\dots \\ & =& \left(2c_2+c_1-c_0\right)+\left(6c_3+2c_2\right)x+\left(12c_4+3c_3+c_2+\frac{1}{2}{c}_1\right)x^2\\ & & +\left(20c_5+4c_4+2c_3+c_2+\frac{1}{6}{c}_1\right)x^3+\dots \end{array}$

This implies that,

$$\begin{gathered} 2c_2+c_1-c_0=0\Rightarrow c_2=\frac{c_0}{2}-\frac{c_1}{2} \\ 6c_3+2c_2=0\Rightarrow c_3=\frac{-c_2}{3} \\ =\frac{-c_0}{6}+\frac{c_1}{6} \\ 12c_4+3c_3+c_2+\frac{1}{2}{c}_1=0\Rightarrow c_4=-\frac{c_3}{4}-\frac{c_2}{12}-\frac{c_1}{24} \\ =\frac{c_0}{24}-\frac{c_1}{24}-\frac{c_0}{24}+\frac{c_y}{24}-\frac{c_1}{24} \\ =\frac{-c_1}{24} \end{gathered}$$

Find the expand the series

Now, we expand the series in (1), which yields

$\begin{array}{rcl}y& =& c_0+c_{1}x+c_2{x}^2+c_3{x}^3+c_4{x}^4+\dots \\ & =& c_0+c_{1}x+\left[\frac{c_0}{2}-\frac{c_1}{2}\right]x^2+\left[\frac{-c_0}{6}+\frac{c_1}{6}\right]x^3+\left[\frac{-c_1}{24}\right]x^4+\dots \\ & =& c_0\underset{y_1}{\underbrace{\left[1+\frac{1}{2}{x}^2-\frac{1}{6}{x}^3+\dots \right]}}+c_1\underset{y_2}{\underbrace{\left[x-\frac{1}{2}{x}^2+\frac{1}{6}{x}^3-\frac{1}{24}{x}^4+\dots \right]}}\\ & & \end{array}$

This is the general solution for the given differential equation. Therefore, the two linearly independent solutions are

$$\begin{gathered} y_1=1+\frac{1}{2}{x}^2-\frac{1}{6}{x}^3+\dots \\ {y}_2=x-\frac{1}{2}{x}^2+\frac{1}{6}{x}^3-\frac{1}{24}{x}^4+\dots \end{gathered}$$