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

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$y = \sec x$

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(A) Given differential equation is $\frac{dy}{dx} = y 	an x$. Separating the variables,we get $\frac{dy}{y} = 	an x \, dx$. Integrating both sides,$\int \frac{dy}{y} = \int 	an x \, dx$. This gives $\ln |y| = \ln |\sec x| + C$,which can be written as $\ln |y| = \ln |C \sec x|$. Thus,$y = C \sec x$ is the general solution. Given that $y = 1$ when $x = 0$,we substitute these values: $1 = C \sec(0)$. Since $\sec(0) = 1$,we have $1 = C 	imes 1$,so $C = 1$. Substituting $C = 1$ into the general solution,we get the particular solution $y = \sec x$.

Find a particular solution satisfying the given condition: $\frac{dy}{dx} = y \tan x$; $y = 1$ when $x = 0$.

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