Question

The general solution of the differential equation \[\frac{dy}{dx} + y \] cot x = cosec x, is

Options

• x + y sin x = C

• x + y cos x = C

• y + x (sin x + cos x) = C

• y sin x = x + C

Answer 1

y sin x = x + C

 

We have, 
\[\frac{dy}{dx} + y \cot x = cosec x\]
\[\frac{dy}{dx} + y\cot x = cosec x\]
\[\text{ Comparing with }\frac{dy}{dx} + Py = Q,\text{ we get }\]
\[P = \cot x \]
\[Q = cosec x\]
Now,
\[I . F . = e^{\int\cot x dx} = e^{log\left( \sin x \right)} \]
\[ = \sin x\]
So, the solution is given by
\[y\sin x = \int\sin x \times\text{ cosec }x dx + C\]
\[ \Rightarrow y \sin x = x + C\]