Solution of differential equation \((\mathrm{dy} / \mathrm{dx})+\mathrm{ay}=\mathrm{e}^{\mathrm{mx}}\) is: (A) \(y=e^{m x}+c \cdot e^{-a x}\) (B) \((\mathrm{a}+\mathrm{m}) \mathrm{y}=\mathrm{e}^{\mathrm{mx}}+\mathrm{c}\) (C) \((a+m) y=e^{m x}+c \cdot e^{-a x}\) (D) \(\mathrm{y} \cdot \mathrm{e}^{\mathrm{ax}}=\mathrm{m} \cdot \mathrm{e}^{\mathrm{mx}}+\mathrm{c}\)

We first find the integrating factor, which is given by the exponential of the integral of the coefficient of y (which is a). To find the integrating factor, we compute: \(I(x) = e^{\int a dx}\)

Now we can calculate the integrating factor: \(I(x) = e^{ax}\)

Next, we multiply both sides of the differential equation by the integrating factor: \(e^{ax} \left(\frac{dy}{dx} + ay\right) = e^{ax}e^{mx}\)

On the left side of the equation, we can use the product rule in reverse to combine the terms into a single derivative: \(\frac{d}{dx}\left(e^{ax}y\right) = e^{ax}e^{mx}\)

Now we integrate both sides of the equation with respect to x: \(\int\frac{d}{dx} \left(e^{ax}y\right)dx = \int e^{ax}e^{mx}dx\) \(e^{ax}y = \int e^{(a+m)x}dx + C\)

Finally, we can solve for y: \(y = e^{-ax}\left(\int e^{(a+m)x}dx + C\right)\) To match one of the answer choices, we split the integral and the constant term: \(y = e^{-ax}\left(\int e^{(a+m)x}dx\right) + e^{-ax}C\) And rewrite the integral in terms of another constant: \(y = e^{-ax}Ae^{(a+m)x} + e^{-ax}C\) Because \(\int e^{(a+m)x}dx = \frac{1}{a+m}e^{(a+m)x} + K\), we can rewrite the equation as: \(y = e^{-ax}\left(\frac{1}{a+m}e^{(a+m)x} + Ce^{-ax}\right)\) Now we compare our solution to the given options: (A) \(y=e^{mx}+c \cdot e^{-ax}\) (B) \((a+m) y=e^{mx}+c\) (C) \((a+m) y=e^{mx}+c \cdot e^{-ax}\) (D) \(y \cdot e^{ax}=m \cdot e^{mx}+c\) In the end, we can see that our solution matches option (C): \((a+m)y = e^{mx} + C \cdot e^{-ax}\)