Find the equation of a curve passing through the origin given that the slope of the tangent to the curve at any point $(x,\;y)$ is equal to the sum of the coordinates of the point.

: We know that the slope of the tangent to the curve is $\cfrac{{dy}}{{dx}}$

We are given that $\cfrac{{dy}}{{dx}} = x + y \Rightarrow \cfrac{{dy}}{{dx}} - y = x$

…(1)
which is a linear equation of the type $\cfrac{{dy}}{{dx}} + Py = Q$

Where, $P = - 1$ and $Q = x$

$\therefore$ ${\rm{I}}{\rm{.F}}{\rm{.}} = {e^{\int P dx}} = {e^{\int { - dx} }} = {e^{ - x}}$

$\therefore$ The solution is $y({\rm{I}}.{\rm{F}}.) = \int Q ({\rm{I}}.{\rm{F}}.)dx + C$

$y \cdot {e^{ - x}} = \int x \cdot {e^{ - x}}dx + C = x \cdot \cfrac{{{e^{ - x}}}}{{ - 1}} - \int {(1)} \cfrac{{{e^{ - x}}}}{{ - 1}}dx + C$

[Integrating by parts]
$\Rightarrow y{e^{ - x}} = - x{e^{ - x}} + \int {{e^{ - x}}} dx + C = - x{e^{ - x}} + \cfrac{{{e^{ - x}}}}{{ - 1}} + C$

$\Rightarrow y = - x - 1 + C{e^x}$

…(2)
Since the curve passes through the origin $(0,0)$,

$\therefore$ $= 0 = - 0 - 1 + C{e^0} \Rightarrow C = 1$

Putting $C = 1$ in (2),

we get $y = - x - 1 + {e^x}$
$\Rightarrow x + y + 1 = {e^x}$,

which is the required equation of the curve.