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(A) The given differential equation is of the form $\frac{dy}{dx} + Py = Q$,where $P = 3$ and $Q = e^{-2x}$.First,we find the integrating factor $(I.F

Vedclass · Accepted Answer

$y = e^{-2x} + Ce^{-3x}$

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(A) The given differential equation is of the form $\frac{dy}{dx} + Py = Q$,where $P = 3$ and $Q = e^{-2x}$.First,we find the integrating factor $(I.F.)$:$I.F. = e^{\int P dx} = e^{\int 3 dx} = e^{3x}$.The general solution is given by:$y(I.F.) = \int (Q 	imes I.F.) dx + C$.Substituting the values:$y e^{3x} = \int (e^{-2x} 	imes e^{3x}) dx + C$$y e^{3x} = \int e^{x} dx + C$$y e^{3x} = e^{x} + C$.Dividing both sides by $e^{3x}$:$y = \frac{e^{x}}{e^{3x}} + \frac{C}{e^{3x}}$$y = e^{-2x} + Ce^{-3x}$.

Find the general solution of the differential equation: $\frac{dy}{dx} + 3y = e^{-2x}$

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