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प्रश्न
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उत्तर
\[\frac{dy}{dx} + 1 = e^{x + y}\] .....(1)
Let x + y = t
\[\Rightarrow 1 + \frac{dy}{dx} = \frac{dt}{dx}\]
Substituting the value of x + y = t and \[1 + \frac{dy}{dx} = \frac{dt}{dx}\] in (1), we get
\[\frac{dt}{dx} = e^t \]
\[ \Rightarrow e^{- t} dt = dx\]
\[ \Rightarrow - e^{- t} = x + C\]
\[ \Rightarrow - e^{- \left( x + y \right)} = x + C ...........\left[ \because t = x + y \right]\]
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