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The Solution of the Differential Equation D Y D X + 1 = E X + Y , is - Mathematics

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प्रश्न

The solution of the differential equation \[\frac{dy}{dx} + 1 = e^{x + y}\], is

पर्याय

  • (x + y) ex + y = 0

  • (x + C) ex + y = 0

  • (x − C) ex + y = 1

  • (x − C) ex + y + 1 =0

MCQ

उत्तर

(x − C) ex + y + 1 = 0

 

We have, 
\[\frac{dy}{dx} + 1 = e^{x + y} \]
\[\text{ Let }x + y = v\]
\[ \Rightarrow 1 + \frac{dy}{dx} = \frac{dv}{dx}\]
\[ \Rightarrow \frac{dy}{dx} + 1 = \frac{dv}{dx}\]
\[ \therefore \frac{dv}{dx} = e^v \]
\[ \Rightarrow e^{- v} dv = dx\]
Integrating both sides, we get
\[ - e^{- v} = x - C\]
\[ \Rightarrow - 1 = e^v \left( x - C \right)\]
\[ \Rightarrow \left( x - C \right) e^{x + y} + 1 = 0\]

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पाठ 22: Differential Equations - MCQ [पृष्ठ १४२]

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आरडी शर्मा Mathematics [English] Class 12
पाठ 22 Differential Equations
MCQ | Q 27 | पृष्ठ १४२

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