Question

solve y'-xy+=0 using power series

Answer 1

Assume a solution of the form \(\large\displaystyle y = \sum_{n=0}^{\infty}c_nx^n\). Then \(\large\displaystyle y^{\prime} = \sum_{n=0}^{\infty}c_nnx^{n-1} = \sum_{n=1}^{\infty}c_n nx^{n-1}\). Plugging this into your differential equation leaves us with
\[\large \begin{aligned}\sum_{n=1}^{\infty}c_n nx^{n-1} - x\sum_{n=0}^{\infty} c_nx^n = 0 &\implies \sum_{n=1}^{\infty}c_n nx^{n-1} - \sum_{n=0}^{\infty}c_n x^{n+1} = 0 \end{aligned}\]
Expanding these two series out and grouping like terms, we get
\[\begin{aligned} &\phantom{=}(c_1+2c_2x+3c_3x^2+4c_4x^3+\cdots)-(c_0 x+c_1x^2+c_2x^3+c_3x^4 +\ldots)\\ &= c_1 + (2c_2-c_0)x+(3c_3-c_1)x^2+(4c_4-c_2)x^3 + (5c_5-c_3)x^4+\ldots = 0\end{aligned}\]
Therefore,\[\large \left\{\begin{aligned}c_1 &= 0\\ 2c_2-c_0 &= 0\\ 3c_3-c_1 &= 0\\ 4c_4 - c_2 &= 0\\ 5c_5 -c_3 &= 0\\ 6c_c -c_4 &= 0\\ &\,\;\vdots \end{aligned}\right. \implies \left\{\begin{aligned}c_1 &= 0\\ c_2 &= \frac{c_0}{2} = \frac{c_0}{2^1\cdot 1!}\\ c_3 &= \frac{c_1}{3} =0\\ c_4 &= \frac{c_2}{4} = \frac{1}{4}\cdot \frac{c_0}{2^1\cdot 1!} = \frac{c_0}{2^2\cdot 2!}\\ c_5 &= \frac{c_3}{5} = 0\\ c_6 &= \frac{c_4}{6} = \frac{1}{6}\cdot\frac{c_0}{2^2\cdot 2!} = \frac{c_0}{2^3\cdot 3!}\\&\,\;\vdots \end{aligned}\right.\]
Thus, we have that all the odd subscript coefficients are zero; Only the even ones survive. We now see that the general solution is
\[\large \begin{aligned}y & =c_0+c_1x+c_2x^2+c_3x^2+c_4x^4+\ldots\\ &= c_0 + \frac{c_0}{2^1\cdot 1!}x^2 + \frac{c_0}{2^2 2!}x^4 + \frac{c_0}{2^3\cdot 3!}x^6 + \ldots \\ &= c_0\sum_{n=0}^{\infty} \frac{x^{2n}}{2^n\cdot n!}\end{aligned} \] I hope this makes sense! :-)