Question

The solution(s) of the differential equation ${\left(\frac{dy}{dx}\right)}^2+2y\cot x\left(\frac{dy}{dx}\right)=y^2$ is/are

• A. $y=\frac{c}{1-\sin x}$, where $c$ is arbitrary constant
• B. $y=\frac{c}{1+\sin x}$, where $c$ is arbitrary constant
• C. $y=\frac{c}{1-\cos x}$, where $c$ is arbitrary constant
• D. $y=\frac{c}{1+\cos x}$, where $c$ is arbitrary constant

Answer 1

Answer: (C), (D)

The correct options are
C $y=\frac{c}{1-\cos x}$, where $c$ is arbitrary constant
D $y=\frac{c}{1+\cos x}$, where $c$ is arbitrary constant

${\left(\frac{dy}{dx}\right)}^2+2y\cot x\left(\frac{dy}{dx}\right)=y^2\Rightarrow {\left(\frac{dy}{dx}\right)}^2+2y\cot x\left(\frac{dy}{dx}\right)-y^2=0$

Assuming $\left(\frac{dy}{dx}\right)=t$, we get
$t^2+2y\cot xt-y^2=0\Rightarrow t=\frac{dy}{dx}=\frac{-2y\cot x\pm \sqrt{4y^2{\cot }^{2}x+4y^2}}{2}\Rightarrow \frac{dy}{dx}=y(-\cot x\pm \text{cosec}x)$

When $\frac{dy}{dx}=y(-\cot x+\text{cosec}x)$, we get
$\Rightarrow \int \frac{dy}{y}=\int (-\cot x+\text{cosec}x)dx\Rightarrow \int \frac{dy}{y}=\int \frac{1-\cos x}{\sin x}dx\Rightarrow \ln |y|=\int \frac{2{\sin }^2\frac{x}{2}}{2\sin \frac{x}{2}\cos \frac{x}{2}}dx\Rightarrow \ln |y|=\int \tan \frac{x}{2}dx\Rightarrow \ln |y|=2\ln \left|\sec \frac{x}{2}\right|+\ln C\Rightarrow y=C_1{\sec }^2\frac{x}{2}\Rightarrow y=\frac{2C_1}{1+\cos x}\therefore y=\frac{c}{1+\cos x}$

When $\frac{dy}{dx}=y(-\cot x-\text{cosec}x)$
$\Rightarrow \int \frac{dy}{y}=-\int (\cot x+\text{cosec}x)dx\Rightarrow \ln y=-\int \frac{1+\cos x}{\sin x}dx\Rightarrow \ln |y|=-\int \frac{2{\cos }^2\frac{x}{2}}{2\sin \frac{x}{2}\cos \frac{x}{2}}+\ln C\Rightarrow \ln |y|=-\int \cot \frac{x}{2}dx+\ln C\Rightarrow \ln |y|=-2\ln \left|\sin \frac{x}{2}\right|+\ln C\Rightarrow y=\frac{2C_1}{2{\sin }^2\frac{x}{2}}\therefore y=\frac{c}{1-\cos x}$