The solution of $\frac{d y}{d x}+1=e^{x+y}$ is
The solution of $\frac{d y}{d x}+1=e^{x+y}$ is
- $e^{-(x+y)}+x+c=0$
- $e^{-(x+y)}-x+c=0$
- $e^{x+y}+x+c=0$
- $e^{x+y}-x+c=0$
Solution
Given, $\quad \frac{d y}{d x}+1=e^{x+y}$
Put
$x+y=z$
$\begin{aligned} & \Rightarrow & 1+\frac{d y}{d x} & =\frac{d z}{d x} \\ & \therefore & \frac{d z}{d x} & =e^z \\ & \Rightarrow & \int e^{-z} d z & =\int d x \\ & \Rightarrow & -e^{-z} & =x+c \\ & \Rightarrow & -e^{-(x+y)} & =x+c \\ & \Rightarrow & x+e^{-(x+y)}+c & =0\end{aligned}$
Asked in: AP EAMCET 2007
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