Answer to Question #133850 in Differential Equations for Aniruddh Thakur

Question

Solve dy/dx^2-y=0 in series

Answer 1

$$y''-y=0$$

Assume that $\displaystyle\sum_{n=0}^\infin a_nx^n$ is a solution. Then you have

$$y'=\displaystyle\sum_{n=1}^\infin na_nx^{n-1},\ y''=\displaystyle\sum_{n=2}^\infin n(n-1)a_nx^{n-2}$$

$$\displaystyle\sum_{n=2}^\infin n(n-1)a_nx^{n-2}=\displaystyle\sum_{n=0}^\infin a_nx^{n}$$

$$\displaystyle\sum_{n=0}^\infin(n+2)(n+1)a_{n+2}x^{n}=\displaystyle\sum_{n=0}^\infin a_nx^{n}$$

By equating coefficients, we have

$$(n+2)(n+1)a_{n+2}=a_n$$

We obtain the recursion formula

$$a_{n+2}=\dfrac{a_n}{(n+2)(n+1)}, n\geq0$$

$a_0, a_1$

$$a_{2}=\dfrac{a_0}{1\cdot2} , \ \ \ \ \ a_{3}=\dfrac{a_1}{2\cdot3}$$

$$a_{4}=\dfrac{a_0}{1\cdot2\cdot3\cdot4} , \ \ \ \ \ a_{5}=\dfrac{a_1}{2\cdot3\cdot4\cdot5}$$

$$\vdots\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \vdots$$

$$a_{2k}=\dfrac{a_0}{(2k)!} , \ \ \ \ \ a_{2k+1}=\dfrac{a_1}{(2k+1)!}$$

Thus, we can represent the general solution as the sum of two series—one for the

even-powered terms with coefficients in terms of $a_0$ and one for the odd-powered

terms with coefficients in terms of $a_1$

$$y=a_0\displaystyle\sum_{k=0}^\infin \dfrac{x^{2k}}{(2k)!}+a_1\displaystyle\sum_{k=0}^\infin \dfrac{x^{2k+1}}{(2k+1)!}$$

The solution has two arbitrary constants, and as we would expect in the

general solution of a second-order differential equation.

$$y=C_1e^x+C_2e^{-x}$$