Question

In Problems 9-14 use an appropriate infinite series method about x = 0to find two solutions of the given differential equation.

${\mathit{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{-}{\mathit{x}}^{\mathbf{2}}{\mathit{y}}^{\mathbf{\text{'}}}\mathbf{+}\mathit{x}\mathit{y}\mathbf{=}\mathbf{0}$

Short answer

Therefore, the solution is

$$\begin{gathered} y_1=x \\ {y}_2=1-\frac{1}{6}{x}^2-\frac{1}{90}{x}^6-\dots \end{gathered}$$

Step-by-step solution

Given Information

The given value is

$y^{\text{'}\text{'}}-x^2{y}^{\text{'}}+xy=0$

Use Differentiate

We were given the next equation

$y^{\text{'}\text{'}}-x^2{y}^{\text{'}}+xy=0----\left(1\right)$

x = 0 is an ordinary point of the DE (1). By Theorem 6.2.1, we can find two linearly independent solutions in the next form

$y=\sum _{n=0}^{\infty }{c}_n{x}^n----\left(2\right)$

The derivations of (2) are

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

Substitute this into (1) to obtain

$0=\sum _{n=2}^{\infty }{c}_{n}n(n-1)x^{n-2}-x^2\sum _{n=1}^{\infty }{c}_{n}n{x}^{n-1}+x\sum _{n=0}^{\infty }{c}_n{x}^n$

$=\sum _{n=2}^{\infty }{c}_{n}n(n-1)x^{n-2}-\sum _{n=1}^{\infty }{c}_{n}n{x}^{n+1}+\sum _{n=0}^{\infty }{c}_n{x}^{n+1}----\left(4\right)$

Now, look at the last equation closer.

Using table

Now, using table, we can rewrite (4) as

$2c_2+\left(6c_3+c_0\right)x+\sum _{n=1}^{\infty }\left[(n+3)(n+2)c_{n+3}-(n-1)c_n\right]x^{n+1}=0----\left(5\right)$

From this, we can see that$c_2=0,c_3=-c_0/6$and

$c_{n+3}=\frac{n-1}{(n+3)(n+2)}{c}_n,n\in \mathrm{\mathbb{N}}----\left(6\right)$

Let c₀= 0 and c₁ = 1 From this, we find

$c_3=c_6=\dots =0$

$c_4=c_7=\dots =0$

$c_5=c_8=\dots =0$

Therefore, one solution is

y₁ = x

Let c₀ = 1 and c₁= 0 From this, we find

$$\begin{gathered} c_3=-\frac{1}{6} \\ {c}_4=c_7=\cdots =0 \\ {c}_5=c_8=\cdots =0 \\ {c}_6=-\frac{1}{90} \end{gathered}$$

Therefore, the second solution is

$y_2=1-\frac{1}{6}{x}^2-\frac{1}{90}{x}^6-\dots$