Question

solve by power series 2y'-y=cosh(x)

Answer 1

2 y' = y + cosh (x) = y + e^x /2 + e^-x  /2
       = y + 1/2 [1 + x + x²/2! + x³/3! + x⁴/4! + x⁵/5! + x⁶/6! +....]
             + 1/2 [1 - x + x²/2! - x³/3! + x⁴/4! - x⁵/5! + x⁶/6! - ... ]
2 y' = y + 1 + x² + x⁴/4! + x⁶/6! + x⁸/8! + ...

Let  y = A + B x + C x² + D x³ + E x⁴ + F x⁵ + G x⁶ + ....     -----(1)
       y' = B + 2 C X + 3 D x² + 4 E x³ + 5 F x⁴ + 6 G x⁵ + ...

2B+4Cx+6D x²+8Ex³+10Fx⁴+12G x⁵ +14Hx⁶...
  = A+Bx+Cx²+Dx³+Ex⁴+Fx⁵+Gx⁶+.... + 1+ x²/2! +x⁴/4! +x⁶/6!+...

Equating coefficients:
   2B=A+1,   4C = B ,  6D = C + 1/2,   8E = D ,  10F = E + 1/4!
  12G = F,    14H = G + 1/6!

=> B = A/2+1/2,  C=A/(2*4) + 1/(2*4),   D=A/(2*4*6) + 5/(2*4*6)
      E = A/(2*4*6*8) + 5/(2*4*6*8)
      F = A/(2*4*6*8*10) + 21/(2*4*6*8*10)

Substitute these values in the equation (1) to get the answer.