Question

The solution of the differential equaton $3e^x\tan ydx+\left(1-e^x\right){\sec }^{2}ydy=0$ is

• A. $\tan y=c{\left(1-e^x\right)}^3$
• B. ${\left(1-e^x\right)}^3\tan y=c$
• C. $\tan y=c\left(1-e^x\right)$
• D. $\left(1-e^x\right)\tan y=c$

Answer 1

The correct option is A $\tan y=c{\left(1-e^x\right)}^3$

$3e^x\tan ydx+(1-e^x){\sec }^{2}ydy=0$

$\frac{3e^x}{1-e^x}dx+\frac{{\sec }^{2}ydy}{\tan y}=0$

Integrating both sides

$-3\log 1-e^x+\log \tan y)=\log e$

$\log \left(\tan y\right)=\log C+3\log (1-e^x)$

$\log \left(\tan y\right)=\log \left(C{(1-e^x)}^3\right)$

$\tan y=C{(1-e^x)}^3$