Question

How can the power series method be used to solve the non-homogeneous equation $\mathit{y}\mathbf{"}\mathbf{-}\mathit{x}\mathit{y}\mathbf{}\mathbf{=}\mathbf{}\mathbf{1}$about the ordinary point x = 0? Of$\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{4}\mathit{x}\mathit{y}\mathbf{\text{'}}\mathbf{-}\mathbf{4}\mathit{y}\mathbf{=}{\mathit{e}}^{\mathbf{x}}$? Carry out your ideas by solving both DEs.

Short answer

For the first differential equation, our solution is:

$y=c_0\left[1+\frac{1}{6}{x}^3+\dots \right]+c_1\left[x+\frac{1}{12}{x}^4+\dots +\right]+\left[\frac{1}{2}{x}^2+\frac{1}{40}{x}^5\right]$

For the second one, we have:

$$\begin{gathered} y=c_0\left[1+2x^2+2x^4+\dots \right]+c_1\left[x+\frac{4}{3}{x}^3+\frac{16}{15}{x}^5\dots \right]+\left[\frac{1}{2}{x}^2+\frac{1}{6}{x}^3+\frac{13}{24}{x}^4+\right. \\ \left.\frac{17}{120}{x}^5+\dots \right] \end{gathered}$$

Step-by-step solution

To Find the differential equation

We have

$y\text{'}\text{'}-xy=1andy\text{'}\text{'}-4x{y}^{\text{'}}]-4y=e^x$

We need to figure out how to solve the given differential equations, using the Power series Method. Well, we have a non-homogeneous second-order differential equation. In the previous problems, we worked on the homogeneous differential equations, where we assumed that the solution has the following form.

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

Where$x_0$is an ordinary point for the differential equation. We used to find the derivatives of y, and substitute them into the differential equation, and equate the coefficients of the variables, raised to the same power for the left and right-hand side of the equation, resulting in a recurrence relation by which we can find the coefficient of the series. The homogeneous case is very simple, as we deal with a zero term on the right-hand side of the equation. Regarding the non-homogeneous case, we deal with constants, other than zero, and variables on the right-hand side. Let's assume that the solution has the following form:

$y=\sum _{n=0}^{\infty }{c}_n{x}^n\dots \left(1\right)where{x}_0=0$

We find the first and second derivatives of (1) as follows;

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

We shift the summation index to n = 1, for y', as the first term of the power series equals zero, and n = 2, for y"

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

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

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

$$\begin{gathered} \sum _{n=2}^{\infty }{c}_{n}n(n-1)x^{n-2}-x\sum _{n=0}^{\infty }{c}_n{x}^n=1 \\ \underset{a_2\to x^0}{\underbrace{\sum _{n=2}^{\infty }{c}_{n}n(n-1)x^{n-2}}}-\underset{a_0\to x^1}{\underbrace{\sum _{n=0}^{\infty }{c}_n{x}^{n+1}}}=1 \end{gathered}$$

Our aim, now, is to make the summation index and the power x , for the two series, in the same phase. We, first have to assure that the first terms, for the two power series, are raised to the same power, which is not the case. Therefore, we do the following;

$2c_2+\underset{a_3\to x^1}{\underbrace{\sum _{n=3}^{}{c}_{n}n(n-1)x^{n-2}}}-\underset{a_0\to x^1}{\underbrace{\sum _{n=0}^{\infty }{c}_n{x}^{n+1}}}=1$

$2c_2+\underset{a_3\to x^1}{\underbrace{\sum _{n=3}^{}{c}_{n}n(n-1)x^{n-2}}}-\underset{a_0\to x^1}{\underbrace{\sum _{n=0}^{\infty }{c}_n{x}^{n+1}}}=1$

Now, we shift the summation index for the first power series. Note that, this wouldn't affect the terms of the series, The shift yields the same terms.

$$\begin{gathered} 2c_2+\sum _{n+3=3}^{\infty }{c}_{n+3}(n+3)\left(\right(n+3)-1)x^{(n+3)-2}-\sum _{n=0}^{\infty }{c}_n{x}^{n+1}=1 \\ 2c_2+\sum _{n=0}^{\infty }{c}_{n+3}(n+3)(n+2)x^{n+1}-\sum _{n=0}^{\infty }{c}_n{x}^{n+1}=1 \\ \underset{coefficientof{x}^0}{\underbrace{2c_2}}+\sum _{n=0}^{\infty }\underset{coefficientof{x}^{n+1}}{\underbrace{\left[c_{n+3}(n+3)(n+2)-c_n\right]}}{x}^{n+1}=\underset{coefficientof{x}^0}{\underbrace{1}}+0x^{n+1} \end{gathered}$$

Comparing and equating the coefficients of variables, which are raised to the same power, yields;

$$\begin{gathered} c_{n+3}(n+3)(n+2)-c_n=0 \\ {c}_{n+3}=\frac{c_n}{(n+3)(n+2)}\dots \left(4\right) \\ And \\ 2c_1=\frac{1}{2} \dots \left(5\right) \\ {c}_2=1 \end{gathered}$$

The second non-homogeneous differential equation

By using (4) and (5), we can find the coefficients of the power series.

$$\begin{gathered} n=0,c_3=\frac{c_0}{6} \\ n=1,c_4=\frac{c_1}{12} \\ n=2,c_5=\frac{c_2}{20}\Rightarrow =\frac{1}{40} \end{gathered}$$

Now we expand the power series in (1).

$$\begin{gathered} y=\sum _{n=0}^{\infty }{c}_n{x}^n \\ =c_0+c_{1}x+c_2{x}^2+c_3{x}^3+c_4{x}^4+\dots \\ =c_0+c_{1}x+c_2{x}^2+c_3{x}^3+c_4{x}^4+\dots \\ =c_0+c_{1}x+\frac{1}{2}{x}^2+\frac{c_0}{6}{x}^3+\frac{c_1}{12}{x}^4+\frac{1}{40}{x}^5+\dots \\ =c_0\left[1+\frac{1}{6}{x}^3+\dots \right]+c_1\left[x+\frac{1}{12}{x}^4+\dots +\right]+\left[\frac{1}{2}{x}^2+\frac{1}{40}{x}^5\right] \end{gathered}$$

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

$$\begin{gathered} \sum _{n=2}^{\infty }{c}_{n}n(n-1)x^{n-2}-4x\sum _{n=1}^{\infty }{c}_{n}n{x}^{n-1}-4\sum _{n=0}^{\infty }{c}_n{x}^n=e^x \\ \underset{a_2\to x^0}{\underbrace{\sum _{n=2}^{\infty }{c}_{n}n(n-1)x^{n-2}}}-\underset{a_1\to x^1}{\underbrace{\sum _{n=1}^{\infty }4c_{n}n{x}^n}}-\underset{a_0\to x^0}{\underbrace{\sum _{n=0}^{\infty }4c_n{x}^n}}=e^x \end{gathered}$$

Our aim, now, is to make the summation index and the power of $x$, for the three series, in the same phase. We, first, have to assure that the first terms, for the three power series, are raised to the same power, which is not the case. Therefore, we do the following:

$$\begin{gathered} \sum c_{n}n(n-1)x^{n-2}-\sum _{n=1}^{\infty }4c_{n}n{x}^n-\sum _{n=0}^{\infty }4c_n{x}^n=e^x \\ 2c_2+\sum _{n=3}^{\infty }{c}_{n}n(n-1)x^{n-2}-\sum _{n=1}^{\infty }4c_{n}n{x}^n-4c_0-\sum _{n=1}^{\infty }4c_n{x}^n\&=\sum _{n=0}^{\infty }\frac{1}{n!}{x}^n \end{gathered}$$

Now, we shift the summation index for the first power series on the left-hand side. And take out the first term of the Maclaurin series. Note that, this wouldn't affect the terms of the series, The shift yields the same terms.

$$\begin{gathered} \left[2c_2-4c_0\right]+\sum _{n+2=3}^{\infty }{c}_{n+2}(n+2)\left(\right(n+2)-1)x^{(n+2)-2}-\sum _{n=1}^{\infty }4c_{n}n{x}^n-\sum _{n=1}^{\infty }4c_n{x}^n=1+\sum _{n=1}^{\infty }\frac{1}{n!}{x}^n \\ 2c_2-4c_0+\sum _{n=1}^{\infty }{c}_{n+2}(n+2)(n+1)x^n-\sum _{n=1}^{\infty }4c_{n}n{x}^n-\sum _{n=1}^{\infty }4c_n{x}^n=1+\sum _{n=1}^{\infty }\frac{1}{n!}{x}^n \\ \underset{coefficientsof{x}^0}{\underbrace{2c_2-4c_0}}+\sum _{n=1}^{\infty }\left[c_{n+2}(n+2)(n+1)-4c_n(n+1)\right]x^n=1+\sum _{n=1}^{\infty }\frac{1}{n!}{x}^n \end{gathered}$$

Final Answer

By comparing and equating the coefficients of variables, which are raised to the same power, yields

$$\begin{gathered} c_{n+2}(n+2)(n+1)-4c_n(n+1)=\frac{1}{n!} \\ {c}_{n+2}(n+2)(n+1)=\frac{1}{n!}+4c_n(n+1)\dots \left(6\right) \\ {c}_{n+2}=\frac{1}{n!(n+2)(n+1)}+\frac{4c_n}{(n+2)} \\ And \\ 2c_2-4c_0=\frac{1+4c_0}{2}\dots \left(5\right) \\ {c}_2=\frac{1}{2}+2c_0\dots \left(6\right) \end{gathered}$$

By comparing and equating the coefficients of variables, which are raised to the same power, yields

$$\begin{gathered} c_{n+2}(n+2)(n+1)-4c_n(n+1)=\frac{1}{n!} \\ {c}_{n+2}(n+2)(n+1)=\frac{1}{n!}+4c_n(n+1)\dots \left(6\right) \\ {c}_{n+2}=\frac{1}{n!(n+2)(n+1)}+\frac{4c_n}{(n+2)} \\ 2c_2-4c_0 \\ {c}_2=\frac{1+4c_0}{2}\dots \left(5\right) \\ =\frac{1}{2}+2c_0\dots \left(6\right) \end{gathered}$$

For the first differential equation, our solution is:

$y=c_0\left[1+\frac{1}{6}{x}^3+\dots \right]+c_1\left[x+\frac{1}{12}{x}^4+\dots +\right]+\left[\frac{1}{2}{x}^2+\frac{1}{40}{x}^5\right]$

For the second one, we have

$y=c_0\left[1+2x^2+2x^4+\dots \right]+c_1\left[x+\frac{4}{3}{x}^3+\frac{16}{15}{x}^5\dots \right]+\left[\frac{1}{2}{x}^2+\frac{1}{6}{x}^3+\frac{13}{24}{x}^4+\right.$

$\left.\frac{17}{120}{x}^5+\dots \right]$