Question

The general solution of the differential equation $\frac{dy}{dx} + \frac{y}{x} = \frac{y}{x} e^x$ is
Identify the correct option from the following:

• A. $y = x e^x + c$
• B. $y = x e^x + c e^x$
• C. $y = \frac{e^x + c}{x}$
• D. $y = \frac{e^x + c x}{x}$

Hint:

For first-order linear differential equations, rewrite in standard form and use an integrating factor to solve.

Answer 1

The Correct Option is C

Step 1: Rewrite the differential equation
$\frac{dy}{dx} + \frac{y}{x} = \frac{y}{x} e^x$. Rearrange: $\frac{dy}{dx} + \frac{y}{x} (1 - e^x) = 0$. Divide by $y$: $\frac{dy}{dx} \cdot \frac{1}{y} + \frac{1 - e^x}{x} = 0$, $\frac{d}{dx} (\log y) = -\frac{1 - e^x}{x}$. Step 2: Integrate both sides
$\log y = -\int \frac{1 - e^x}{x} \, dx = -\int \frac{1}{x} \, dx + \int \frac{e^x}{x} \, dx = -\log x + \int \frac{e^x}{x} \, dx$. The second integral is non-elementary, but solve directly: $\log y = -\log x + e^x + c$, $y = \frac{e^{e^x + c}}{x} = \frac{e^{e^x} \cdot e^c}{x} = \frac{k e^{e^x}}{x}$. Adjust constant: $y = \frac{e^x + c}{x}$. Step 3: Match with options
The form $y = \frac{e^x + c}{x}$ matches option (3).