Since
y(0) = 1 and y'(0) = 2, the solution of the initial value problem can
be written as
the following power series: y(x) = 1 + 2x + a_{2}x^{2}
+ a_{3}x^{3}
+ a_{4}x^{4}
+ … .

Differentiate:
y '(x) = 2 + 2a_{2}x
+ 3a_{3}x^{2}
+ 4a_{4}x^{3
}+…
and y ''(x) = 2a_{2}
+ 6a_{3}x
+ 12a_{4}x^{2}
+ … .

Substitute
the expansion of the function y(x) and its second derivative into the
differential equation: