Differential equations with laplace

[dumbengineer]

Homework Statement

D^2 x -2x = 2sin2t x(0) = 0 , x'(0) = 1

Homework Equations

D^2 x = s^2ƒ[x] - sx(0) - x'(o)

The Attempt at a Solution

s^2ƒ[x] - sx(0) - x'(o) - 2ƒ[x] = 2sin2t

s^2ƒ[x] - sx(0) - x'(o) - 2ƒ[x] = 4 / s^2 + 4

ƒ[x] (s^2 + 1) + 1 = 4 / s^2 +4


ƒ[x] (s^2 + 1) = 4 / s^2 +4 - 1/s

i need help from here, i just keep messing up from this point and on, any help is greatly appreciated thanks !

 

[LCKurtz]

Homework Statement

D^2 x -2x = 2sin2t x(0) = 0 , x'(0) = 1

Homework Equations

D^2 x = s^2ƒ[x] - sx(0) - x'(o)

The Attempt at a Solution

s^2ƒ[x] - sx(0) - x'(o) - 2ƒ[x] = 2sin2t

s^2ƒ[x] - sx(0) - x'(o) - 2ƒ[x] = 4 / s^2 + 4

ƒ[x] (s^2 + 1) + 1 = 4 / s^2 +4

Perhaps, being careful with your arithmetic and parentheses you mean
f[x](s²-2) -1 = 4/(s²+4)

Your careless use of parentheses may be causing additional errors in your next steps, which you didn't show.