The general solution of the equation frac{dy} | vedclass

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(A) The given differential equation is of the form $\frac{dy}{dx} + P(x)y = Q(x)$,where $P(x) = \frac{1}{x}$ and $Q(x) = \frac{e^x}{x}$.First,we find

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$y = \frac{e^x + c}{x}$

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(A) The given differential equation is of the form $\frac{dy}{dx} + P(x)y = Q(x)$,where $P(x) = \frac{1}{x}$ and $Q(x) = \frac{e^x}{x}$.First,we find the integrating factor $(IF)$:$IF = e^{\int P(x) dx} = e^{\int \frac{1}{x} dx} = e^{\ln|x|} = x$.The general solution is given by $y \cdot (IF) = \int Q(x) \cdot (IF) dx + c$.Substituting the values:$y \cdot x = \int \frac{e^x}{x} \cdot x dx + c$$xy = \int e^x dx + c$$xy = e^x + c$$y = \frac{e^x + c}{x}$.

The general solution of the equation $\frac{dy}{dx} + \frac{1}{x}y = \frac{1}{x}e^x$ is

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