y' = e^3x (3sin4x+4cos4x)
y"' = e^3x (24cos4x-7sin4x)
so,
y"-6y'+25y = e^3x(24cos4x-7sin4x-6(3sin4x+4cos4x)+25sin4x) = 0
turning points where y'=0
3sin4x+4cos4x = 0
x = odd multiples of pi/2 +/- arctan(4/3)
I think
The only asymptote is where e^3x = 0, or y=0
if y=e^(3x) sin 4x. Show that (d^2y/dx^2)-(6dy/dx)+25y=0
(b) for the curve with equation y=(x^2+1)/(x^2-4), find
(i) the cordinates of the turning point(s).
(ii) the equation of the asymptotes
1 answer