Question

Find the particular solution of e^dy/dx = x + 1, given that y = 3, when x = 0.

Answer 1

We have, 
\[ e^\frac{dy}{dx} = x + 1\]
\[ \Rightarrow \frac{dy}{dx} = \log \left( x + 1 \right)\]
\[ \Rightarrow dy = \log \left( x + 1 \right) dx\]
Integrating both sides, we get 
\[\int dy = \int\log \left( x + 1 \right) dx\]
\[ \Rightarrow y = \log \left( x + 1 \right)\int1 dx - \int\left[ \frac{d}{dx}\left\{ \log \left( x + 1 \right) \right\}\int1 dx \right]dx\]
\[ \Rightarrow y = x \log \left( x + 1 \right) - \int\frac{1}{x + 1} \times x dx\]
\[ \Rightarrow y = x \log \left( x + 1 \right) - \int\left( 1 - \frac{1}{x + 1} \right) dx\]
\[ \Rightarrow y = x \log \left( x + 1 \right) - \int dx + \int\frac{1}{x + 1}dx\]
\[ \Rightarrow y = x \log \left( x + 1 \right) - x + \log \left| x + 1 \right| + C\]
\[ \Rightarrow y = \left( x + 1 \right) \log \left| x + 1 \right| - x + C . . . . . (1)\]
It is given that at x = 0 and y = 3 . 
Substituing the values of x and y in (1), we get 
\[C = 3\]
Therefore, substituting the value of C in (1), we get
\[y = \left( x + 1 \right) \log \left| x + 1 \right| - x + 3\]
\[\text{ Hence, }y = \left( x + 1 \right) \log \left| x + 1 \right| - x + 3 \text{ is the required solution . }\]