The equation of the curve for which the slope of the tangent at any point is given by (x + y + 1)((d y/d x))=1 is (where, c is an arbitrary constant)

Question

The equation of the curve for which the slope of the tangent at any point is given by (x + y + 1)((d y/d x))=1 is (where, c is an arbitrary constant) (A) xy=ex-c (B) xy=cey+2 (C) x=cey-y-2 (D) x=ey+y-c

Tardigrade Admin · Accepted Answer

The given equation is (d x/d y)+x(- 1)=y+1 Integrating Factor =e- displaystyle ∫ d y=e- y Thus, the curve is x ⋅ (e- y) = displaystyle ∫ (y + 1) e- y d y ⇒ xe- y=-(y + 1)e- y-e- y+c or x=cey-y-2

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

$x y = e^{x} - c$

$x y = c e^{y} + 2$

$x = c e^{y} - y - 2$

$x = e^{y} + y - c$

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

Q. The equation of the curve for which the slope of the tangent at any point is given by (x+y+1)(dxdy​)=1 is (where, c is an arbitrary constant)

xy=ex−c

xy=cey+2

x=cey−y−2

x=ey+y−c