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These are the steps that i was told to follow however at the end i can only come up with a general solution with unknown variables please help thank you

 Sep 16, 2019

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 #1
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\(\dfrac{dy}{dx}+y = \sin(2x)\\ a=1,~b=1,~f(x)=\sin(2x)\\ y_{cf} = Ae^{-\frac b a x}=A e^{-x}\\ \text{try $y_{pt} = B\sin(2x) + C\cos(2x)$}\\ 2B\cos(2x) - 2C \sin(2x) + B\sin(2x)+C\cos(2x) = \sin(2x)\\ 2B+C = 0\\ B-2C = 1\\ 5B = 1\\ B=\dfrac 1 5,~C = -\dfrac 2 5\\ y = y_{cf}+y_{pt} = A e^{-x} + \dfrac 1 2 \left(\sin(2x)-2\cos(2x)\right)\)

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 Sep 16, 2019
 #1
avatar+6046 
+1
Best Answer

\(\dfrac{dy}{dx}+y = \sin(2x)\\ a=1,~b=1,~f(x)=\sin(2x)\\ y_{cf} = Ae^{-\frac b a x}=A e^{-x}\\ \text{try $y_{pt} = B\sin(2x) + C\cos(2x)$}\\ 2B\cos(2x) - 2C \sin(2x) + B\sin(2x)+C\cos(2x) = \sin(2x)\\ 2B+C = 0\\ B-2C = 1\\ 5B = 1\\ B=\dfrac 1 5,~C = -\dfrac 2 5\\ y = y_{cf}+y_{pt} = A e^{-x} + \dfrac 1 2 \left(\sin(2x)-2\cos(2x)\right)\)

Rom Sep 16, 2019

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