Question

The solution of $x \, dy - y \, dx + x^2 \, e^x \, dx = 0$ is :

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

Answer 1

Answer: (A)

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

Answer 2

The diff. e can be written as: $\frac{dy}{dx} - \frac{y}{x} = - xe^{x}$ $I.f. =e^{-\int \frac{1}{x} dx} = e^{-\log x} = \frac{1}{x}$ Thus, the solution is; $y \frac{1}{x} = \int -xe^{x}. \frac{1}{x} dx +c$ $\Rightarrow \frac{y}{x} = - e^{x} +c$ $\Rightarrow \frac{y}{x} + e^{x} = c$