(a). \( y^{\prime}+3 y=2 e^{-x} \quad \operatorname{Integ} \)
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To solve the differential equation \( y' + 3y = 2e^{-x} \), we’ll use the integrating factor method. The integrating factor is given by \( e^{\int 3 \, dx} = e^{3x} \). Multiplying the entire equation by this factor yields: \[ e^{3x}y' + 3e^{3x}y = 2e^{2x} \] This simplifies the left side to: \[ \frac{d}{dx}(e^{3x}y) = 2e^{2x} \] Integrating both sides gives: \[ e^{3x}y = \frac{2}{3} e^{2x} + C \] Where \( C \) is the constant of integration. Finally, solving for \( y \): \[ y = \frac{2}{3} e^{-x} + Ce^{-3x} \] Now you've got a general solution!