मराठी

X D Y D X + X Cos 2 ( Y X ) = Y - Mathematics

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प्रश्न

\[x\frac{dy}{dx} + x \cos^2 \left( \frac{y}{x} \right) = y\]

बेरीज
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उत्तर

We have,

\[x\frac{dy}{dx} + x \cos^2 \left( \frac{y}{x} \right) = y\]

\[\frac{dy}{dx} + \cos^2 \left( \frac{y}{x} \right) = \frac{y}{x}\]

\[ \Rightarrow \frac{dy}{dx} = \frac{y}{x} - \cos^2 \left( \frac{y}{x} \right)\]

Putting `y = vx,` we get

\[\frac{dy}{dx} = v + x\frac{dv}{dx}\]

\[ \therefore v + x\frac{dv}{dx} = v - \cos^2 \left( v \right)\]

\[ \Rightarrow x\frac{dv}{dx} = - \cos^2 v\]

\[ \Rightarrow \sec^2 v\ dv = - \frac{1}{x}dx\]

Integrating both sides, we get

\[\int sec^2 v\ dv = - \int\frac{1}{x}dx\]

\[ \Rightarrow \tan v = - \log x + C\]

\[ \Rightarrow \tan \frac{y}{x} = - \log x + C\]

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पाठ 22: Differential Equations - Revision Exercise [पृष्ठ १४६]

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आरडी शर्मा Mathematics [English] Class 12
पाठ 22 Differential Equations
Revision Exercise | Q 54 | पृष्ठ १४६

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