Question

If $y\left(\frac{\pi }{2}\right)=e,$ then the solution of $y^{\prime }\sin x=y\ln y$ is

• A. $|y|=e^{\text{cosec}x+\cot x}$
• B. $y=e^{|\text{cosec}x-\cot x|}$
• C. $y=e^{\cos x-\text{cosec}x}$
• D. $y=e^{\sec x+\tan x}$

Answer 1

The correct option is B $y=e^{|\text{cosec}x-\cot x|}$

$\frac{dy}{dx}\sin x=y\ln y\Rightarrow \int \frac{dy}{y\ln y}=\int \text{cosec}xdx$

$\Rightarrow \ln |\ln y|=\ln |\text{cosec}x-\cot x|+\ln |C|$
$\Rightarrow |\ln y|=|C(\text{cosec}x-\cot x)|$

at $x=\frac{\pi }{2},y=e;$ so $\pm \ln \left(e\right)=C(1-0)\Rightarrow C=\pm 1$

$\Rightarrow y=e^{|(\text{cosec}x-\cot x)|}$