Q120 / 160 · TS EAMCET 2007 · Mathematics

The solution of $\frac{d y}{d x}+1=e^{x+y}$ is

  • Correct answer
    A
    $e^{-(x+y)}+x+c=0$
  • B
    $e^{-(x+y)}-x+c=0$
  • C
    $e^{x+y}+x+c=0$
  • D
    $e^{x+y}-x+c=0$

Solution

Given, $\quad \frac{d y}{d x}+1=e^{x+y}$ Put $x+y=z$ $\begin{aligned} & \Rightarrow & 1+\frac{d y}{d x} & =\frac{d z}{d x} \\ & \therefore & \frac{d z}{d x} & =e^z \\ & \Rightarrow & \int e^{-z} d z & =\int d x \\ & \Rightarrow & -e^{-z} & =x+c \\ & \Rightarrow & -e^{-(x+y)} & =x+c \\ & \Rightarrow & x+e^{-(x+y)}+c & =0\end{aligned}$

Hence, the correct answer is option: 'A'.

Relevant Questions