Eliminate the arbitrary constant and obtain the differential equation satisfied by it
$y = 2 x + c e^{x}$

y^{'} - y = 2 (1 - x)

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y^{'} - y = 2 (3 - x)

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y^{'} - y = 2 (1 - 3 x)

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y^{'} - y = 3 (1 - x)

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Solution

The correct option is A $y^{'} - y = 2 (1 - x)$
$y = 2 x + c e^{x}$ ...(1)
Differentiating it w.r.t x, we get
$y^{'} = 2 + c e^{x}$ ...(2)
Subtracting (1) from (2), we get
$y^{'} - y = 2 + c e^{x} - 2 x + c e^{x} = 2 - 2 x \Rightarrow y^{'} - y = 2 (1 - x)$

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