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D Y D X = ( E X + 1 ) Y

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Question

\[\frac{dy}{dx} = \left( e^x + 1 \right) y\]
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Solution

We have,
\[\frac{dy}{dx} = \left( e^x + 1 \right) y\]
\[ \Rightarrow \frac{1}{y}dy = \left( e^x + 1 \right) dx\]
Integrating both sides, we get 
\[\int\frac{1}{y}dy = \int\left( e^x + 1 \right) dx\]
\[ \Rightarrow \log \left| y \right| = e^x + x + C\]
\[\text{ Hence, }\log \left| y \right| = e^x + x +\text{ C is the required solution .} \]

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Chapter 21: Differential Equations - Exercise 22.07 [Page 55]

APPEARS IN

R.D. Sharma Mathematics Volume 1 and 2 [English] Class 12
Chapter 21 Differential Equations
Exercise 22.07 | Q 3 | Page 55

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