The solution of frac{dy}{dx} + frac{1}{3}y = | vedclass

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(B) The given differential equation is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = \frac{1}{3}$ and $Q = 1$.The int

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$y = 3 + ce^{-x/3}$

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(B) The given differential equation is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = \frac{1}{3}$ and $Q = 1$.The integrating factor $(IF)$ is given by $IF = e^{\int P dx} = e^{\int \frac{1}{3} dx} = e^{x/3}$.The general solution is $y 	imes (IF) = \int (Q 	imes IF) dx + c$.Substituting the values,we get $y 	imes e^{x/3} = \int (1 	imes e^{x/3}) dx + c$.Integrating the right side,$y 	imes e^{x/3} = 3e^{x/3} + c$.Dividing both sides by $e^{x/3}$,we get $y = 3 + ce^{-x/3}$.

The solution of $\frac{dy}{dx} + \frac{1}{3}y = 1$ is

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