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 ?
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$y = Ae^{ax} \cos (bx) + Be^
$y = Ae^{ax} \cos (bx) + Be^{ax} \sin (bx)$ ← 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
In reply to $y = Ae^{ax} \cos (bx) + Be^ by Jhun Vert
Thanks for this, this helps a lot :)