Answer to Question #202284 in Differential Equations for Rajeev

Question

Using power series method to solve the initial value problem

(x − )1 y′′ + yx ′ + y = 0 , y )0( = 2 , y′ )0( = −1 .

Answer 1

given, the differential equation

(x − 1) y′′ +x y′ + y = 0,

y (0)= 2 , y′(0)= −1.

Let the solution be $y=\sum c_nx^n \newline y'=\sum nc_nx^{n-1} \newline y''=\sum n(n-1)c_nx^{n-2} \newline \text{substitute the above value in the given equation, we get}\newline (x-1)\sum_{n=2}^\infty n(n-1)c_nx^{n-2}+x\sum_{n=1}^\infty nc_nx^{n-1}+\sum_{n=0}^\infty c_nx^n=0 \newline \sum_{n=2}^\infty n(n-1)c_nx^{n-1}-\sum_{n=2}^\infty n(n-1)c_nx^{n-2} -\sum_{n=1}^\infty nc_nx^{n}+\sum_{n=0}^\infty c_nx^n=0 \newline \sum_{n=2}^\infty n(n+1)c_{n+1}x^{n}-\sum_{n=2}^\infty (n+1)(n+2)c_{n+2}x^{n} -\sum_{n=1}^\infty nc_nx^{n}+\sum_{n=0}^\infty c_nx^n=0 \newline 2c_2-c_1+\sum_{n=0}^\infty n(n+1)c_{n+1}x^{n}-\sum_{n=0}^\infty (n+1)(n+2)c_{n+2}x^{n} -\sum_{n=0}^\infty nc_nx^{n}+\sum_{n=0}^\infty c_nx^n=0 \newline \text{We get,}\newline 2c_2-c_1=0 \text{and the recurrence relation is,} c_{n+2}=\frac{n(n+1)c_{n+1}+(1-n)c_{n}}{(n+1)(n+2)}.\newline \text{Thus, the solution is given by }\newline y=c_0y1+c_1y_2.\\ Let c_0=1, c_1=0, \implies y=y_1.\\ Then, c_2=1/2.\\ Therefore, y_1=1+\frac{1}{2}x^2+... .\\ Similarly,\newline c_0=0, c_1=1, \implies y=y_2.\newline Then, c_2=0, c_3=0.\newline Therefore, y_2=x$