MATH HELP, Power Series Solution for differential equation

Question

Find the general solution of the equation

y' = x³ - 2xy

with power series

Answer 1

−APPROACH ONE

Let y = ∑_n=0 a_n xⁿ . Then y' = ∑_n=1 n a_n x^n-1.

Hence - x³ + ∑_n=1 n a_n x^n-1.+ ∑_n=0 2a_n xⁿ⁺¹ = 0

then - x³ + a₁ + 2a₂x +3a₃x²+4a₄x³ + ∑_n=5 n a_n x^n-1+2a₀x +2a₁x²+ 2a₂x³ + ∑_n=3 2a_n xⁿ⁺¹ = 0

a₁ + 2a₂x +2a₀x +3a₃x²+2a₁x² -x³+4a₄x³ +2a₂x³ + ∑_n=5 n a_n x^n-1+ ∑_n=3 2a_n xⁿ⁺¹ = 0

a₁ +[ 2a₂ +2a₀] x+[3a₃+2a₁ ] x² [-1+4a₄ +2a₂]x³ + ∑_n=5 n a_n x^n-1+ ∑_n=3 2a_n xⁿ⁺¹ = 0

a₁ +[ 2a₂ +2a₀] x+[3a₃+2a₁ ] x² [-1+4a₄ +2a₂]x³ + ∑_n=3 [n+2] a_n+2 xⁿ⁺¹+ ∑_n=3 2a_n xⁿ⁺¹ = 0

a₁ +[ 2a₂ +2a₀] x+[3a₃+2a₁ ] x² [-1+4a₄ +2a₂]x³ + ∑_n=3{ [n+2] a_n+2 + 2a_n} xⁿ⁺¹ = 0

Then equating all coefficients to zero we get

a₁=0

2a₂ +2a₀=0

3a₃+2a₁ =0

-1+4a₄ +2a₂ =0

[n+2] a_n+2 + 2a_n=0 ∀ n ≥ 3

a₀= k _, a₁=0 , a₂= -k , a₃=0 , a₄= (1+2k)/4 ,

and from the recursive formula a_n+2 = - 2a_n/[n+2]=0 ∀ n ≥ 3

a₅= 0, a₆=(-1/3) a₄ , a₇=0 , a₈= ( -1/4) a₆ ....

Then y = ∑_n=0 a_n xⁿ = a₀ + a₁x +a₂x² + a₃x³ +.....

= k - kx² + [(1+2k)/4] x⁴ - [ (1+2k)/12] x⁶ + [ (1+2k)/60] x⁸ -.... now if we let (1+2k)/4 = λ

= k - kx² + (2λx⁴) / ( 2!) - (2λx⁶) / ( 3!) + (2λx⁸) / ( 4!)-.... and if 2 λ = c

= c-1/2 - [c-1/2]x² + cx⁴/ (2!) - cx⁶/ (3!) + cx⁸/ (4!) -...

= -1/2 +x² /2 +c -cx²+ cx⁴/ 2! -cx⁶/ 3! + cx⁸/4! -...

-1/2 +x² /2 +c[1 -x²+ x⁴/ 2! -x⁶/ 3! + x⁸/4! -...] =

= -1/2 +x² /2 + c e^(−x²)

By the way the differential equation is a linear of first order and the above result can be verified easily