Find the equation of a curve passing through origin, if the slope of the tangent to the curve at any point $(x,y)$ is equal to the square of the difference of the abcissa and ordinate of the point.

Slope of tangent to the curve $= \frac{{dy}}{{dx}}$

and difference of abscissa and ordinate $= x - y$

According to the question, $\frac{{dy}}{{dx}} = {(x - y)^2}$
……(i)
Put $x - y = z$

$\Rightarrow$ $1 - \frac{{dy}}{{dx}} = \frac{{dz}}{{dx}}$

$\Rightarrow$ $\frac{{dy}}{{dx}} = 1 - \frac{{dz}}{{dx}}$

On substituting these values in Eq. (i),

we get
$1 - \frac{{dz}}{{dx}} = {z^2}$

$\Rightarrow$ $1 - {z^2} = \frac{{dz}}{{dx}}$

$\Rightarrow$ $dx = \frac{{dz}}{{1 - {z^2}}}$

On integrating both sides,

we get
$\int d x = \int {\frac{{dz}}{{1 - {z^2}}}}$

$\Rightarrow$ $x = \frac{1}{2}\log \left| {\frac{{1 + z}}{{1 - z}}} \right| + C$

$\Rightarrow$ $tx = \frac{1}{2}\log \left| {\frac{{1 + x - y}}{{1 - x + y}}} \right| + C$
……(ii)

Since, the curve passes through the origin.

$\therefore \quad 0 = \frac{1}{2}\log \left| {\frac{{1 + 0 - 0}}{{1 - 0 + 0}}} \right| + C$

$\Rightarrow$ $C = 0$

On substituting the value of $C$ in Eq.

(ii), we get
$x = \frac{1}{2}\log \left| {\frac{{1 + x - y}}{{1 - x + y}}} \right|$

$\Rightarrow$ $2x = \log \left| {\frac{{1 + x - y}}{{1 - x + y}}} \right|$

$\Rightarrow$ ${e^{2x}} = \left| {\frac{{1 + x - y}}{{1 - x + y}}} \right|$

$\Rightarrow$ $(1 - x + y){e^{2x}} = 1 + x - y$