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प्रश्न
The general solution of the differential equation \[\frac{dy}{dx} + y \] cot x = cosec x, is
विकल्प
x + y sin x = C
x + y cos x = C
y + x (sin x + cos x) = C
y sin x = x + C
उत्तर
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
\[y\sin x = \int\sin x \times\text{ cosec }x dx + C\]
\[ \Rightarrow y \sin x = x + C\]
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