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Question Number 123417 by sdfg last updated on 25/Nov/20

	Answered by TANMAY PANACEA last updated on 25/Nov/20

	$\frac{d y}{d x} - 2 y = 3 e^{x}$ $e^{- 2 x} \frac{d y}{d x} - 2 {y e}^{- 2 x} = 3 e^{- x} [i n t r e g a t i n g f a c t o r = e^{\int p d x} = e^{\int - 2 d x}$ $w h e n e q n i s \frac{d y}{d x} + p y = q]$ $\frac{d}{d x} ({y e}^{- 2 x}) = 3 e^{- x}$ $\int d ({y e}^{- 2 x}) = \int 3 e^{- x}$ ${y e}^{- 2 x} = \frac{3 e^{- x}}{- 1} + c$ $0 \times e^{0} = \frac{3 e^{- 0}}{- 1} + c \to c = 3$ ${y e}^{- 2 x} + 3 e^{- x} = 3$
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