Question

The integrating factor of the differential equation 4xdy-e^-2y dy + dx = 0 is

• A. e^-2y
• B. \(e^{2x^{2}}\)
• C. e^4y
• D. e^-4y
• E. x⁴

Answer 1

The Correct Option is C

The given differential equation is: \[ 4x \, dy - e^{-2y} \, dy + dx = 0 \] Rearrange the equation as: \[ (4x - e^{-2y}) \, dy = -dx \] We see that this is a first-order linear differential equation in the form: \[ M(x, y) \, dy + N(x, y) \, dx = 0 \] where \( M(x, y) = 4x - e^{-2y} \) and \( N(x, y) = -1 \). The integrating factor \( \mu(y) \) is given by: \[ \mu(y) = e^{\int P(y) \, dy} \] where \( P(y) \) is the coefficient of \( dy \) divided by \( N(x, y) \). In this case: \[ P(y) = \frac{4x - e^{-2y}}{-1} = e^{-2y} \] Thus, the integrating factor is: \[ \mu(y) = e^{\int e^{-2y} \, dy} = e^{4y} \]

The correct option is (C) : e^4y

Answer 2

We are given the differential equation: 

\( 4x\,dy - e^{-2y}\,dy + dx = 0 \)

Rearranged as:
\[ dx + (4x - e^{-2y})\,dy = 0 \]
or
\[ \frac{dx}{dy} = - (4x - e^{-2y}) \]

This is a linear differential equation in \( x \):
\[ \frac{dx}{dy} + 4x = e^{-2y} \]

This is of the standard form:
\[ \frac{dx}{dy} + P(y)\,x = Q(y) \]
where \( P(y) = 4 \) and \( Q(y) = e^{-2y} \)

The integrating factor (IF) is:
\[ \textbf{IF} = e^{\int 4\,dy} = e^{4y} \]

Correct answer: \( \boxed{e^{4y}} \)