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Question

(A) The given differential equation is: $e^{x} 	an y \, dx + (1 - e^{x}) \sec^{2} y \, dy = 0$. Rearranging the terms,we get: $(1 - e^{x}) \sec^{2} y

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$	an y = C(1 - e^{x})$

Vedclass · Answer

(A) The given differential equation is: $e^{x} 	an y \, dx + (1 - e^{x}) \sec^{2} y \, dy = 0$. Rearranging the terms,we get: $(1 - e^{x}) \sec^{2} y \, dy = -e^{x} 	an y \, dx$. Separating the variables,we get: $\frac{\sec^{2} y}{	an y} \, dy = \frac{-e^{x}}{1 - e^{x}} \, dx$. Integrating both sides: $\int \frac{\sec^{2} y}{	an y} \, dy = \int \frac{-e^{x}}{1 - e^{x}} \, dx$. For the left side,let $	an y = u$,then $\sec^{2} y \, dy = du$. Thus,$\int \frac{du}{u} = \ln|u| = \ln|	an y|$. For the right side,let $1 - e^{x} = t$,then $-e^{x} \, dx = dt$. Thus,$\int \frac{dt}{t} = \ln|t| = \ln|1 - e^{x}|$. Combining these,we get: $\ln|	an y| = \ln|1 - e^{x}| + \ln|C|$. Using logarithmic properties: $\ln|	an y| = \ln|C(1 - e^{x})|$. Taking the exponential of both sides,we obtain the general solution: $	an y = C(1 - e^{x})$.

Find the general solution of the differential equation: $e^{x} \tan y \, dx + (1 - e^{x}) \sec^{2} y \, dy = 0$.

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