The particular solution of the differential e | vedclass

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(B) Given the differential equation: $\frac{dy}{dx} = y 	an x$. Separating the variables,we get: $\frac{dy}{y} = 	an x \, dx$. Integrating both side

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$y = \sec x$

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(B) Given the differential equation: $\frac{dy}{dx} = y 	an x$. Separating the variables,we get: $\frac{dy}{y} = 	an x \, dx$. Integrating both sides: $\int \frac{dy}{y} = \int 	an x \, dx$. This gives: $\ln |y| = \ln |\sec x| + C$. Exponentiating both sides: $|y| = e^{\ln |\sec x| + C} = e^C \cdot |\sec x|$. Let $e^C = k$,so $y = k \sec x$. Using the initial condition $y(0) = 1$: $1 = k \sec(0) \implies 1 = k(1) \implies k = 1$. Substituting $k = 1$ back into the equation,we get the particular solution: $y = \sec x$.

The particular solution of the differential equation $\frac{dy}{dx} = y \tan x$ with the initial condition $y(0) = 1$ is:

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