Differentiate y = e^x * [C sin (2x) + D cos (2x)] twice
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Differentiate y = e^x * [C sin (2x) + D cos (2x)] twice

[From: ] [author: ] [Date: 11-05-25] [Hit: ]
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 +vuwill give me the second differentiation?Im sorry if this completely confusing but I am abit lost on where to go,......
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

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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
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keywords: Differentiate,cos,sin,twice,Differentiate y = e^x * [C sin (2x) + D cos (2x)] twice
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