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

Vedclass · Accepted Answer

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

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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 $e^{\int P dx} = e^{\int \frac{1}{3} dx} = e^{x/3}$.Multiplying both sides of the equation by the $IF$,we get:$e^{x/3} \frac{dy}{dx} + \frac{1}{3} e^{x/3} y = e^{x/3}$This can be written as:$\frac{d}{dx} (y \cdot e^{x/3}) = e^{x/3}$Integrating both sides with respect to $x$:$y \cdot e^{x/3} = \int e^{x/3} dx + c$$y \cdot e^{x/3} = 3e^{x/3} + c$Dividing by $e^{x/3}$:$y = 3 + ce^{-x/3}$

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

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