The particular solution of $\dfrac{dy}{dx} + y an x = \sec x$ given $…

Question

The particular solution of $\dfrac{dy}{dx} + y	an x = \sec x$ given $y(0) = 1$ is:

Accepted Answer

A. $y\sec x = 	an x + 1$

Linear: $P = 	an x$, IF = $e^{\int 	an x\,dx} = e^{\log\sec x} = \sec x$. So $y\sec x = \int \sec x \cdot \sec x\,dx + c = 	an x + c$. At $x=0, y=1$: $1\cdot 1 = 0 + c$ ⇒ $c = 1$. Hence $y\sec x = 	an x + 1$.

Answer

$y\sec x = \sec x + 1$

Answer

$y\sec x = 	an x - 1$

Answer

$y = \sin x + \cos x$

The particular solution of $\dfrac{dy}{dx} + y\tan x = \sec x$ given $y(0) = 1$ is: