The general solution of the differential equa | vedclass

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Question

(D) Given the differential equation: $(\sec x + 	an x) \frac{dy}{dx} + (\sec^2 x + \sec x 	an x) y = 1$ Divide by $(\sec x + 	an x)$: $\frac{dy}{dx

Vedclass · Accepted Answer

$(\sec x + 	an x) y = x + c$

Vedclass · Answer

(D) Given the differential equation: $(\sec x + 	an x) \frac{dy}{dx} + (\sec^2 x + \sec x 	an x) y = 1$ Divide by $(\sec x + 	an x)$: $\frac{dy}{dx} + \sec x \cdot y = \frac{1}{\sec x + 	an x}$ Since $\frac{1}{\sec x + 	an x} = \sec x - 	an x$, the equation becomes: $\frac{dy}{dx} + (\sec x) y = \sec x - 	an x$ This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$, where $P = \sec x$ and $Q = \sec x - 	an x$. Integrating Factor $(IF)$ $= e^{\int P dx} = e^{\int \sec x dx} = e^{\ln|\sec x + 	an x|} = \sec x + 	an x$. The general solution is $y \cdot (IF) = \int Q \cdot (IF) dx + c$. $y(\sec x + 	an x) = \int (\sec x - 	an x)(\sec x + 	an x) dx + c$ $y(\sec x + 	an x) = \int (\sec^2 x - 	an^2 x) dx + c$ Since $\sec^2 x - 	an^2 x = 1$: $y(\sec x + 	an x) = \int 1 dx + c$ $y(\sec x + 	an x) = x + c$.

The general solution of the differential equation $(\sec x + \tan x) \frac{dy}{dx} + (\sec^2 x + \sec x \tan x) y = 1$ is

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