y=3e^(2x)cos(2x-3) verify that d^2y/dx^2-4dy/dx+8y=0
plz help me i tried all i could but it become too complicated for me
here
set u=3e^(2x) v=cos(2x-3)
du/dx=6e^(2x)
i used chain rule
dv/dx=-2sin(2x-3)
dy/dx=-3e^(2x)sin(2x-3)+cos(2x-3)6e^(2x)
d^2y/dx^2
now i did
lndy/dx=-ln3e^(2x)+lnsin(2x-3)+lncos(2x-3)+ln6e^(2x)
d^2y/dx^2.1/dy/dx=-6e^(2x)/(3e^2(x)+2cos(2x-3)/sin(2x-3)-2sin(2x-3)/cos(2x-3)+12e^(2x)
so whats up any 1 see it too muc
see related questions below.
To verify the given equation, let's start by finding the second derivative of y (d^2y/dx^2) using the values of u and v you defined:
dy/dx = -3e^(2x)sin(2x-3) + 6e^(2x)cos(2x-3)
du/dx = 6e^(2x)
dv/dx = -2sin(2x-3)
Now, let's find d^2y/dx^2:
d^2y/dx^2 = d/dx (-3e^(2x)sin(2x-3) + 6e^(2x)cos(2x-3))
= (-3e^(2x)cos(2x-3) - 3e^(2x)sin(2x-3)(-2sin(2x-3)) + 6e^(2x)(-2sin(2x-3))) + (6e^(2x)(-2cos(2x-3)) - 6e^(2x)sin(2x-3)(2cos(2x-3)))
Simplifying this expression step by step:
d^2y/dx^2 = (-3e^(2x)cos(2x-3) + 6e^(2x)sin(2x-3)sin(2x-3) + 12e^(2x)sin(2x-3)) + (-12e^(2x)cos(2x-3) - 12e^(2x)sin(2x-3)cos(2x-3))
Now, let's collect like terms:
d^2y/dx^2 = (12e^(2x)sin(2x-3) - 12e^(2x)cos(2x-3)) + (-3e^(2x)cos(2x-3) + 6e^(2x)sin(2x-3)sin(2x-3) + 12e^(2x)sin(2x-3))
Rearranging the terms:
d^2y/dx^2 = [12e^(2x)sin(2x-3) - 3e^(2x)cos(2x-3)] + [6e^(2x)sin(2x-3)sin(2x-3) - 12e^(2x)cos(2x-3) + 12e^(2x)sin(2x-3)]
Now, let's rewrite the expression in terms of u and v:
d^2y/dx^2 = uv - 3u + v^2 - 2v + 2v - 2uv
= uv - 2uv - 3u + v^2 - 2v + 2v
= -u(v + 2) - 3u + v^2
Now, substituting back the values of u and v:
d^2y/dx^2 = -[3e^(2x)] * [(cos(2x-3)) + 2)] - 3 * [3e^(2x)] + [cos(2x-3)]^2
Simplifying further:
d^2y/dx^2 = -3e^(2x)(cos(2x-3) + 2) - 9e^(2x) + cos^2(2x-3)
Now, let's express the original equation:
d^2y/dx^2 - 4(dy/dx) + 8y = 0
Plugging in the values for d^2y/dx^2, dy/dx, and y:
[-3e^(2x)(cos(2x-3) + 2) - 9e^(2x) + cos^2(2x-3)] - 4[-3e^(2x)sin(2x-3) + 6e^(2x)cos(2x-3)] + 8[3e^(2x)cos(2x-3)]
= 0
Simplifying this equation should yield 0 if it is true.