Question

The solution of
$3e^x{\cos }^{2}ydx+(1-e^x)\cot ydy=0$

• A. $\tan y=c(e^x-1)$
• B. ${\tan }^{2}y=(e^x-1)c$
• C. $\cot y=c(e^x-1{)}^2$
• D. $\tan y=c(e^x-1{)}^3$

Answer 1

Answer: (D)

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

$3e^x{\cos }^{2}ydx=(e^x-1)\cot ydy$

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

Put $e^x-1=t$ ; $tany=k$

$\Rightarrow e^{x}dx=dt$ ; ${\sec }^{2}ydy=dk$

$\Rightarrow \int \frac{3dt}{t}+logc=\int \frac{dk}{k}$

$\Rightarrow log{t}^{3}c=logk$

$\Rightarrow k=c{t}^3$

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