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(C) The given differential equation is $\frac{dy}{dx} + y = e^x$.This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$, where $P

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$2ye^x = e^{2x} + C$

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(C) The given differential equation is $\frac{dy}{dx} + y = e^x$.This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$, where $P = 1$ and $Q = e^x$.The integrating factor $(IF)$ is given by $IF = e^{\int P dx} = e^{\int 1 dx} = e^x$.The general solution is given by $y \cdot (IF) = \int Q \cdot (IF) dx + C$.Substituting the values, we get $y \cdot e^x = \int e^x \cdot e^x dx + C$.$y e^x = \int e^{2x} dx + C$.$y e^x = \frac{e^{2x}}{2} + C$.Multiplying both sides by $2$, we get $2ye^x = e^{2x} + 2C$.Letting $2C = C_1$, we have $2ye^x = e^{2x} + C_1$.

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

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