$\cfrac{x}{a} + \cfrac{y}{b} = 1$

.: We have, $\cfrac{x}{a} + \cfrac{y}{b} = 1$ …(1)

Differentiating (1) w.r.t. $x$,

we get $\cfrac{1}{a} + \cfrac{1}{b}y' = 0$

…(2)
Again differentiating (2) w.r.t. $x$,

we get
$0 + \cfrac{1}{b}{y^n} = 0 \Rightarrow {y^n} = 0$

which is the required differential equation.