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The General Solution of the Differential Equation D Y D X + Y Cot X = Cosec X, is - Mathematics

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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

MCQ
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Solution

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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Chapter 22: Differential Equations - MCQ [Page 140]

APPEARS IN

RD Sharma Mathematics [English] Class 12
Chapter 22 Differential Equations
MCQ | Q 6 | Page 140

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