Hi, I am trying to solve a differential equation, and I have already obtained the complementary function but I am trying to find the particular integral, how would I differentiate the above equation twice to find d^2*y/d*x^2 ?
my answer so far to find dy/dx is:
dy/dx = e^x { [2C cos (2x)  2D sin (2x)] + [C sin (2x) + D cos (2x)] }
if this is correct, would I need to use the chain rule again to find the d^2*y/d*x^2, thereby letting e^x = u and { [2C cos (2x)  2D sin (2x)] + [C sin (2x) + D cos (2x)] } = v ???
so that using the product rule: uv' +vu' will give me the second differentiation?
I'm sorry if this completely confusing but I am abit lost on where to go, ANY help would be great!! THANKS
my answer so far to find dy/dx is:
dy/dx = e^x { [2C cos (2x)  2D sin (2x)] + [C sin (2x) + D cos (2x)] }
if this is correct, would I need to use the chain rule again to find the d^2*y/d*x^2, thereby letting e^x = u and { [2C cos (2x)  2D sin (2x)] + [C sin (2x) + D cos (2x)] } = v ???
so that using the product rule: uv' +vu' will give me the second differentiation?
I'm sorry if this completely confusing but I am abit lost on where to go, ANY help would be great!! THANKS

y = e^x * [C sin (2x) + D cos (2x)] twice
Divide both side by e^x
y*e^(x) = [C sin (2x) + D cos (2x)]...............(i)
differentiate both side
(dy/dx)*(e^(x) y{e^(x)} = 2[C.cos(2x)  D sin(2x)]..........(ii)
differentiate again
[{d^2y/dx^2)}*{e^(x)}  (dy/dx){e^(x)}] [(dy/dx){e^(x)}{(y){e^(x)}] = 4[C.sin(2x) +D cos(2x)]..........(iii)
{e^(x)[y" 2y'+y] = 4y
y" 2y' +4y e^x = 0.................Ans
Divide both side by e^x
y*e^(x) = [C sin (2x) + D cos (2x)]...............(i)
differentiate both side
(dy/dx)*(e^(x) y{e^(x)} = 2[C.cos(2x)  D sin(2x)]..........(ii)
differentiate again
[{d^2y/dx^2)}*{e^(x)}  (dy/dx){e^(x)}] [(dy/dx){e^(x)}{(y){e^(x)}] = 4[C.sin(2x) +D cos(2x)]..........(iii)
{e^(x)[y" 2y'+y] = 4y
y" 2y' +4y e^x = 0.................Ans