Search any question & find its solution
Question:
Answered & Verified by Expert
The solution of $\frac{d y}{d x}+1=e^{x+y}$ is
Options:
Solution:
2147 Upvotes
Verified Answer
The correct answer is:
$e^{-(x+y)}+x+c=0$
Given, $\quad \frac{d y}{d x}+1=e^{x+y}$
Put
$x+y=z$
$\begin{aligned} & \Rightarrow & 1+\frac{d y}{d x} & =\frac{d z}{d x} \\ & \therefore & \frac{d z}{d x} & =e^z \\ & \Rightarrow & \int e^{-z} d z & =\int d x \\ & \Rightarrow & -e^{-z} & =x+c \\ & \Rightarrow & -e^{-(x+y)} & =x+c \\ & \Rightarrow & x+e^{-(x+y)}+c & =0\end{aligned}$
Put
$x+y=z$
$\begin{aligned} & \Rightarrow & 1+\frac{d y}{d x} & =\frac{d z}{d x} \\ & \therefore & \frac{d z}{d x} & =e^z \\ & \Rightarrow & \int e^{-z} d z & =\int d x \\ & \Rightarrow & -e^{-z} & =x+c \\ & \Rightarrow & -e^{-(x+y)} & =x+c \\ & \Rightarrow & x+e^{-(x+y)}+c & =0\end{aligned}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.