Question

The integrating factor of the differential equation \[ \frac{dy}{dx} + y \tan x - \sec x = 0 \quad \text{is:} \]

• A. $-\cos x$
• B. $\sec x$
• C. $\log \sec x$
• D. $e^{\sec x}$

Hint:

For linear first-order differential equations of the form $\frac{dy}{dx} + P(x)y = Q(x)$, the integrating factor is always $e^{\int P(x) \, dx}$. In this case, $P(x) = \tan x$, and the integrating factor is $\sec x$.

Answer 1

The Correct Option is B

The differential equation is given as:\[\frac{dy}{dx} + y \tan x - \sec x = 0.\]This can be rewritten as:\[\frac{dy}{dx} + y \tan x = \sec x.\]This is a first-order linear differential equation in the standard form:\[\frac{dy}{dx} + P(x)y = Q(x),\]with $P(x) = \tan x$ and $Q(x) = \sec x$. The solution employs an integrating factor, defined as:\[\mu(x) = e^{\int P(x) \, dx}.\]Substituting $P(x) = \tan x$, we calculate:\[\mu(x) = e^{\int \tan x \, dx}.\]The integral of $\tan x$ is:\[\int \tan x \, dx = \log \sec x.\]Therefore, the integrating factor is:\[\mu(x) = e^{\log \sec x} = \sec x.\]The integrating factor is $\sec x$.