.: We have, a + b = 1 …(1) Differentiating (1) w.r.t. x , we get a + b y' = 0 …(2) Again differentiating (2) w.r.t. x , we get 0 + b y n = 0 = 0 which is the required differential equation.

Differential Equations — Class 12 Maths Solution

ncert exercise SA NCERT Ex.9.3,Q.1,Page 391

Question

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

Step-by-step Solution

.: 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.

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NCERT & Exemplar solution for CBSE Class 12 Mathematics, Differential Equations. Curated by Sachin Sharma. Free for all students.