Question

The solution of the differential equation \( e^x y dx + e^x dy + xdx = 0 \) is:

• A. \( e^x + yx^2 = c \)
• B. \( 2ye^x + x^2 = c \)
• C. \( ye^x + x^2 e^y = c \)
• D. \( e^x + x e^y = c \)

Hint:

Check whether the given differential equation is exact before proceeding with integration.

Answer 1

The Correct Option is B

To solve the differential equation \( e^x y dx + e^x dy + x dx = 0 \), we can follow these steps: Given Differential Equation: \[ e^x y dx + e^x dy + x dx = 0 \] Step 1: Rearrange the Equation Combine like terms: \[ e^x y dx + x dx + e^x dy = 0 \] Factor out \( dx \): \[ (e^x y + x) dx + e^x dy = 0 \] Step 2: Check for Exactness A differential equation of the form \( M dx + N dy = 0 \) is exact if \( \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \). Here, \( M = e^x y + x \) and \( N = e^x \). Compute the partial derivatives: \[ \frac{\partial M}{\partial y} = e^x \] \[ \frac{\partial N}{\partial x} = e^x \] Since \( \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \), the equation is exact. Step 3: Find the Potential Function Integrate \( M \) with respect to \( x \): \[ \int (e^x y + x) dx = y e^x + \frac{x^2}{2} + h(y) \] Differentiate this result with respect to \( y \) and set it equal to \( N \): \[ \frac{\partial}{\partial y} \left( y e^x + \frac{x^2}{2} + h(y) \right) = e^x + h'(y) \] Set this equal to \( N \): \[ e^x + h'(y) = e^x \] Thus, \( h'(y) = 0 \), so \( h(y) \) is a constant. Step 4: Write the General Solution The potential function is: \[ y e^x + \frac{x^2}{2} = C \] Multiply through by 2 to eliminate the fraction: \[ 2 y e^x + x^2 = 2C \] Let \( c = 2C \), then: \[ 2 y e^x + x^2 = c \] Final Answer: \[ \boxed{2 y e^x + x^2 = c} \] This corresponds to option (2).