Question

The equation of the curve through $(0,\frac{\pi }{4})$ satisfying the differential equation. $e^x\tan ydx+(1+e^x){\sec }^{2}ydy=0$ is given by

• A. $(1+e^x)\tan y=2$
• B. $1+e^x=2\tan y$
• C. $1+e^x=2\sec y$
• D. $(1+e^x)\tan y=1$

Answer 1

Answer: (D)

$(1+e^x)\tan y=1$

$e^x\tan ydx+(1+e^x){\sec }^{2}ydy=0$

$(1+e^x){\sec }^{2}ydy=-e^x\tan ydx$

$\frac{{\sec }^{2}y}{\tan y}dy=\frac{-e^x}{1+e^x}dx$

substitute $u=\tan y\to du={\sec }^{2}ydy$ and $v=1+e^x\to dv=e^{x}dx$

$\int \frac{1}{u}du=\int \frac{-1}{v}dv$

$\ln u=-\ln v$

$\ln \left(\tan y\right)=-\ln (1+e^x)$

$\ln \left(\tan y\right)+\ln (1+e^x)=0$

$\ln (1+e^x)\tan y=0$

$(1+e^x)\tan y=1$