Elimination of arbitrary constants: $y = Ae^{ax} \cos (bx) + Be^{ax} \sin (bx)$

Im having a difficulty solving this problem y = Aeaxcos(bx)+Beaxsin(bx) where a and b are parameters. Can anyone show me how to solve this ?

Jhun Vert's picture

$y = Ae^{ax} \cos (bx) + Be^{ax} \sin (bx)$   ←   Equation (1)

$y' = A \Big[ -be^{ax} \sin (bx) + ae^{ax} \cos (bx) \Big] + B \Big[ be^{ax} \cos (bx) + ae^{ax} \sin (bx) \Big]$

$y' = a \Big[ Ae^{ax} \cos (bx) + Be^{ax} \sin (bx) \Big] + bBe^{ax} \cos (bx) - bAe^{ax} \sin (bx)$

$y' = ay + bBe^{ax} \cos (bx) - bAe^{ax} \sin (bx)$   ←   Equation (2)

$y'' = ay' + bB \Big[ -be^{ax} \sin (bx) + ae^{ax} \cos (bx) \Big] - bA \Big[ be^{ax} \cos (bx) + ae^{ax} \sin (bx) \Big]$

$y'' = ay' - b^2 \Big[ Ae^{ax} \cos (bx) + Be^{ax} \sin (bx) \Big] + abBe^{ax} \cos (bx) - abAe^{ax} \sin (bx)$

$y'' = ay' - b^2y + abBe^{ax} \cos (bx) - abAe^{ax} \sin (bx)$   ←   Equation (3)

Equation (3) - a × Equation (2)
$y'' - ay' = (ay' - b^2y) - a^2y$

$y'' - 2ay' + (a^2 + b^2)y = 0$           answer

Thanks for this, this helps a lot :)

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