Question #278822

Find two power series solutions of the given differential equation about the ordinary point x = 0. y'' - y = 0

Expert's answer

y=∑n=0∞anxny=\displaystyle\sum_{n=0}^{\infin} a_nx^n


y′′=∑n=2∞n(n−1)anxn−2y''=\displaystyle\sum_{n=2}^{\infin} n(n-1)a_nx^{n-2}


∑n=2∞n(n−1)anxn−2−∑n=0∞anxn=0\displaystyle\sum_{n=2}^{\infin} n(n-1)a_nx^{n-2}-\displaystyle\sum_{n=0}^{\infin} a_nx^n=0


∑n=0∞(n+1)(n+2)an+2xn−∑n=0∞anxn=0\displaystyle\sum_{n=0}^{\infin} (n+1)(n+2)a_{n+2}x^{n}-\displaystyle\sum_{n=0}^{\infin} a_nx^n=0


∑n=0∞[(n+1)(n+2)an+2−an]xn\displaystyle\sum_{n=0}^{\infin} [(n+1)(n+2)a_{n+2}-a_n]x^{n}


an+2=an(n+1)(n+2)a_{n+2}=\frac{a_n}{(n+1)(n+2)}


a2k=a0(2k)!,a2k+1=a1(2k+1)!a_{2k}=\frac{a_0}{(2k)!},a_{2k+1}=\frac{a_1}{(2k+1)!}


y(x)=a0∑k=0∞x2k(2k)!+a1∑k=0∞x2k+1(2k+1)!y(x)=a_0\displaystyle\sum_{k=0}^{\infin}\frac{x^{2k}}{(2k)!}+a_1\displaystyle\sum_{k=0}^{\infin}\frac{x^{2k+1}}{(2k+1)!}


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