मराठी

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

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प्रश्न

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

पर्याय

  • 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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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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पाठ 21: Differential Equations - MCQ [पृष्ठ १४०]

APPEARS IN

आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
पाठ 21 Differential Equations
MCQ | Q 6 | पृष्ठ १४०

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