Question

solve y'=x+y by power series method of solution

Answer 1

Let, solution is

$$y=\sum_{n=0}^{\infty}a_nx^n$$

Thus,

$$y'=\sum_{n=1}^{\infty}a_nx^{n-1}$$

Hence, on simplification the differential equation

$$y'=y+x$$

We get,

$$a_0=a_1,a_2=\frac{a_0+1}{2}, a_3=\frac{a_0+1}{6}$$

and

$$\frac{a_n}{a_{n-1}}=\frac{1}{n}, \forall n\geq 3$$

Thus,

$$a_n=\frac{6}{n!}, \forall n\geq 3$$

Hence, solution will be,

$$y=a_0+a_1x+(\frac{a_0+1}{2})x^2+(a_0+1)\sum_{n=3}^{\infty}\frac{1}{n!}\\ \implies y=a_0+a_1x+(\frac{a_0+1}{2})x^2+(a_0+1)(e^x-1-x-\frac{1}{2}x^2)\\ y=(a_0+1)e^x-x-1$$