Solve the differential equation $e^{dy/dx} = x + 1$.
Also find the particular solution when the initial conditions are $y \left(0\right) = 3$, $x > -1$.
Given: $e^{dy/dx} = x + 1$
i.e. $\log \left|e^{dy/dx}\right| = \log \left|x + 1\right|$
i.e. $\dfrac{dy}{dx} = \log \left|x + 1\right|$
i.e. $dy = \log \left|x + 1\right| \; dx$ $\;\;\; \cdots \; (1)$
Integrating equation $(1)$ gives
$\displaystyle \int dy = \int \log \left|x + 1\right| \; dx$ $\;\;\; \cdots \; (2)$
$\begin{aligned}
\text{Now, } \int \log \left|x + 1\right| \; dx & = \log \left|x + 1\right| \int dx - \int \left\{\int dx \times \dfrac{d}{dx} \left(\log \left|x + 1\right|\right) \right\} \; dx \\\\
& = x \log \left|x + 1\right| - \int x \times \dfrac{1}{x +1} \; dx \\\\
& = x \log \left|x + 1\right| - \left[\int \dfrac{x + 1}{x + 1} \; dx - \int \dfrac{dx}{x + 1}\right] \\\\
& = x \log \left|x + 1\right| - \int dx + \int \dfrac{dx}{x + 1} \\\\
& = x \log \left|x + 1\right| - x + \log \left|x + 1\right| + c_1 \\\\
& = \left(x + 1\right) \log \left|x + 1\right| - x + c_1 \;\;\; \cdots \; (3)
\end{aligned}$
and $\displaystyle \int dy = y + c_2$ $\;\;\; \cdots \; (4)$
$\therefore$ $\;$ In view of equations $(3)$ and $(4)$, equation $(2)$ becomes
$y + c_2 = \left(x + 1\right) \log \left|x + 1\right| - x + c_1$
i.e. $y = \left(x + 1\right) \log \left|x + 1\right| - x + c$ $\;\;\; \cdots \; (5)$ $\;$ where $c = c_1 - c_2$
Equation $(5)$ is the general solution of the given differential equation.
The initial condition is $y \left(0\right) = 3$
i.e. when $x = 0$, $y = 3$
Substituting the values of x and y in equation $(5)$ gives
$3 = \left(0 + 1\right) \log \left|0 + 1\right| - 0 + c$
i.e. $3 = \log \left|1\right| + c$ $\implies$ $c = 3$ $\;\;\; \cdots \; (6)$
Substituting the value of c from equation $(6)$ in equation $(5)$ gives
$y = \left(x + 1\right) \log \left|x + 1\right| - x + 3$ $\;\;\; \cdots \; (7)$
Equation $(7)$ is the particular solution of the given differential equation when $y \left(0\right) = 3$.