1. ## General Solution

Can someone help me with this please

D^2Y
+2DY +2Y=SinX
DX^2 DX

Find the G.S. given that the CF is e^-x(C1Cosx+C2SinX)

2. Hello, Mathsnewbie!

What methods have you been taught?
And there is a typo . . .

$\frac{d^2y}{dx^2} + 2\frac{dy}{dx} + 2y \:=\:\sin x$

Find the G.S. given that the CF is: $e^{-x}(C_1\cos{\color{red}2}x+C_2\sin{\color{red}2}x)$
Undetermined Coefficients

Let: $y \:=\:A\sin x + B\cos x \quad\Rightarrow\quad 2y \:=\:2A\sin x + 2B\cos x$ .[1]

. . $\frac{dy}{dx} \:=\: A\cos x -B\sin x\quad\Rightarrow\quad 2\frac{dy}{dx} \:=\:\text{-}2B\sin x + 2A\cos x$ .[2]

. . $\frac{d^2y}{dx^2} \:=\:\text{-}A\sin x - B\cos x$ .[3]

Add [1], [2], [3]: . $\frac{d^2y}{dx^2} + 2\frac{dy}{dx} + 2y \:=\:(A-2B)\sin x + (2A+B)\cos x \;=\;\sin x$

And we have: . $\begin{array}{ccc}A - 2B &=& 1 \\ 2A + B &=& 0 \end{array}$

Solve the system: . $A \,=\,\tfrac{1}{5},\;B \,=\,\text{-}\tfrac{2}{5}$

. . Hence, the particular solution is: . $y \:=\:\tfrac{1}{5}\sin x - \tfrac{2}{5}\cos x$

The General Solution is: . $y \;=\;e^{-x}\left(C_1\cos2x + C_2\sin2x\right) + \tfrac{1}{5}\sin x - \tfrac{2}{5}\cos x$