Question

The equation of the curve for which the slope of the tangent at any point is given by $\left(x + y + 1\right)\left(\frac{d y}{d x}\right)=1$ is (where, $c$ is an arbitrary constant)

• A. $xy=e^{x}-c$
• B. $xy=ce^{y}+2$
• C. $x=ce^{y}-y-2$
• D. $x=e^{y}+y-c$

Answer 1

Answer: (C)

The given equation is $\frac{d x}{d y}+x\left(- 1\right)=y+1$
Integrating Factor $=e^{- \displaystyle \int d y}=e^{- y}$
Thus, the curve is $x \cdot \left(e^{- y}\right) = \displaystyle \int \left(y + 1\right) e^{- y} d y$
$\Rightarrow xe^{- y}=-\left(y + 1\right)e^{- y}-e^{- y}+c$
or $x=ce^{y}-y-2$