Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSDifferential Equations5 Marks
Question
Find the general solution $ x \frac { d y } { d x } + y - x + x y \cot x = 0 \ (x \neq 0)$.
✓
Answer
Given differential equation is, $ x \frac { d y } { d x } + y - x + x y \cot x = 0$ Above equation can be written as $ x \frac { d y } { d x } + y (1 + x \cot x ) = x$ On dividing both sides with x, we get $ \frac { d y } { d x } + y \left( \frac { 1 + x \cot x } { x } \right) = 1$ $ \Rightarrow \frac { d y } { d x } + y \left( \frac { 1 } { x } + \cot x \right) = 1$ which is a linear differential equation of the form $ \frac { d y } { d x } + P y = Q$, where $ P = \frac { 1 } { x } + \cot x$ and $ Q = 1.$ we know that , $\mathrm { IF } = e ^ { \int { Pdx } } = e ^ { \int \left( \frac { 1 } { x } + \cot x \right) d x } = e ^ { \log | x | + \log \sin x }$ $= e ^ { \log | x \sin x | } [ \because \log m + \log n = \log m n ]$ $ \Rightarrow$ IF = x sin x $y \times { IF } = \int ( Q \times {I F } ) d x + C$ $\therefore \quad y \times x \sin x = \int 1 \times x \sin x d x + C$ $ \Rightarrow yx\sin x = \int {\mathop x\limits_I } \mathop {\sin }\limits_{II} xdx + C$ $\Rightarrow y \cdot x \sin x = x \int \sin x d x$ $- \int \left( \frac { d } { d x } ( x ) \int \sin x d x \right) d x + C$ [using integration by parts] $\Rightarrow y x \sin x = - x \cos x - \int 1 ( - \cos x ) d x + C$ $\Rightarrow y x \sin x = - x \cos x + \int \cos x d x + C$ $ \Rightarrow y x \sin x = - x \cos x + \sin x + C$ On dividing both sides by x sin x, we get $y = \frac { - x \cos x + \sin x + C } { x \sin x }$ $ \therefore y = - \cot x + \frac { 1 } { x } + \frac { C } { x \sin x }$ which is the required solution.
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