The solution of the differential equation 3e^ | vedclass

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Question

(A) Given differential equation: $3e^x 	an y \, dx + (1 - e^x) \sec^2 y \, dy = 0$. Rearranging the terms to separate the variables: $(1 - e^x) \sec^

Vedclass · Accepted Answer

$	an y = c(1 - e^x)^3$

Vedclass · Answer

(A) Given differential equation: $3e^x 	an y \, dx + (1 - e^x) \sec^2 y \, dy = 0$. Rearranging the terms to separate the variables: $(1 - e^x) \sec^2 y \, dy = -3e^x 	an y \, dx$. Dividing both sides by $(1 - e^x) 	an y$: $\frac{\sec^2 y}{	an y} \, dy = -3 \frac{e^x}{1 - e^x} \, dx$. Integrating both sides: $\int \frac{\sec^2 y}{	an y} \, dy = -3 \int \frac{e^x}{1 - e^x} \, dx$. Let $u = 	an y$,then $du = \sec^2 y \, dy$. Let $v = 1 - e^x$,then $dv = -e^x \, dx$. The integral becomes: $\int \frac{1}{u} \, du = 3 \int \frac{1}{v} \, dv$. $\ln|u| = 3 \ln|v| + \ln|c|$. $\ln|	an y| = \ln|c(1 - e^x)^3|$. Therefore,$	an y = c(1 - e^x)^3$.

The solution of the differential equation $3e^x \tan y \, dx + (1 - e^x) \sec^2 y \, dy = 0$ is

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