Solve the differential equation $\dfrac{dy}{dx} = e^{x + y}$
The given differential equation is $\;$ $\dfrac{dy}{dx} = e^{x + y}$ $\;\;\; \cdots \; (1)$
Let $x + y = u$ $\;\;\; \cdots \; (2a)$
Differentiating equation $(2a)$ gives
$1 + \dfrac{dy}{dx} = \dfrac{du}{dx}$
i.e. $\dfrac{dy}{dx} = \dfrac{du}{dx} -1$ $\;\;\; \cdots \; (2b)$
$\therefore$ $\;$ In view of equations $(2a)$ and $(2b)$, equation $(1)$ becomes
$\dfrac{du}{dx} - 1 = e^u$
i.e. $\dfrac{du}{dx} = e^u + 1$
i.e. $\dfrac{du}{e^u + 1} = dx$ $\;\;\; \cdots \; (3)$
Integrating equation $(3)$ gives
$\displaystyle \int \dfrac{du}{e^u + 1} = \int dx$ $\;\;\; \cdots \; (4)$
Consider $\displaystyle \int \dfrac{du}{e^u + 1}$ $\;\;\; \cdots \; (5)$
Let $e^u + 1 = t$ $\;\;\; \cdots \; (5a)$
Differentiating equation $(5a)$ gives
$e^u \; du = dt$
i.e. $du = \dfrac{dt}{e^u} = \dfrac{dt}{t - 1}$ $\;\;\; \cdots \; (5b)$ $\;\;$ [by equation $(5a)$]
$\therefore$ $\;$ In view of equations $(5a)$ and $(5b)$, equation $(4)$ becomes
$\displaystyle \int \dfrac{dt / t - 1}{t} = \int dx$
i.e. $\displaystyle \int \dfrac{dt}{t \left(t - 1\right)} = \int dx$ $\;\;\; \cdots \; (6)$
Let $\dfrac{1}{t \left(t - 1\right)} = \dfrac{A}{t} + \dfrac{B}{t - 1}$ $\;\;\; \cdots \; (7)$
i.e. $1 = A \left(t - 1\right) + B \; t$ $\;\;\; \cdots \; (7a)$
Putting $t = 0$ in equation $(7a)$ gives
$1 = - A$ $\implies$ $A = -1$ $\;\;\; \cdots \; (7b)$
Putting $t = 1$ in equation $(7a)$ gives
$B = 1$ $\;\;\; \cdots \; (7c)$
$\therefore$ $\;$ In view of equations $(7b)$ and $(7c)$, equation $(7)$ becomes
$\dfrac{1}{t \left(t - 1\right)} = \dfrac{-1}{t} + \dfrac{1}{t - 1}$ $\;\;\; \cdots \; (8)$
Integrating equation $(8)$ gives
$\displaystyle \int \dfrac{dt}{t \left(t - 1\right)} = \int \dfrac{-dt}{t} + \int \dfrac{dt}{t - 1}$ $\;\;\; \cdots \; (8a)$
$\therefore$ $\;$ In view of equation $(8a)$, equation $(6)$ can be written as
$\displaystyle \int \dfrac{-dt}{t} + \int \dfrac{dt}{t - 1} = \int dx$
i.e. $- \log \left|t\right| + \log \left|t - 1\right| + \log \left|c\right| = x$
i.e. $\log \left|\dfrac{c \left(t - 1\right)}{t}\right| = x$
i.e. $\dfrac{c \left(t - 1\right)}{t} = e^x$ $\;\;\; \cdots \; (9)$
Substituting the value of $t$ from equation $(5a)$ in equation $(9)$ gives
$\dfrac{c \left(e^u + 1 - 1\right)}{e^u + 1} = e^x$
i.e. $\dfrac{c \; e^u}{e^u + 1} = e^x$ $\;\;\; \cdots \; (10)$
In view of equation $(2a)$, equation $(10)$ becomes
$\dfrac{c \; e^{x+y}}{e^{x+y} + 1} = e^x$
i.e. $\dfrac{c \; e^y}{e^{x+y} + 1} = 1$
i.e. $c \; e^y = e^{x+y} + 1$ $\;\;\; \cdots \; (11)$
Equation $(11)$ is the general solution of the given differential equation.