You are correct that
To solve for and , try substitution:
let g(x) = e^(-4x) * (Dsin3x + Ecos3x)
where D and E are constants. Find values D and E if
g'(x) = e^(-4x) * sin3x
Here's my work:
*must find the derivative and then equate to e^(-4x) * sin3x
g'(x) = [e^(-4x) * -4* (D*cos3x + E*sin3x)] + [e^(-4x) * (3D*cos3x - 3E*sin3x)]
simplified, and factored
g'(x) = (e^(-4x)) * [(-4D-3E)(sin3x) + (3d-4E)(cos3x)]
so now we equate to the derivative from the equation
(e^(-4x)) * [(-4D-3E)(sin3x) + (3D-4E)(cos3x)] = e^(-4x) * sin3x
we can assume that
(3D-4E) = 0 and (-4D-3E) = 1
THIS is where i'm stuck. I know i need to find the value of one in order to find the value of the other constant value, but HOW?!
in previous problems we would have a third constant that normally worked out to be set as something, but in this we dont.
Either i messed up big time in my work or I'm missing something quite obvious.
Either way please please help
Thanks,
Brittany