Differential Equations — Class 12 Maths Solution

ncert misc SA NCERT Misc.,Q.3,Page 420
Question

Form the differential equation representing the family of curves given by
${(x - a)^2} + 2{y^2} = {a^2}$
where ${\rm{a}}$ is an arbitrary constant.

Step-by-step Solution

${(x - a)^2} + 2{y^2} = {a^2}$
$\Rightarrow$ ${x^2} + {a^2} - 2ax + 2{y^2} = {a^2}$

$\Rightarrow$ $2{y^2} = 2ax - {x^2}$

……(1)
Differentiating with respect to ${\rm{x}}$,

we get:
$2y\frac{{dy}}{{dx}} = \frac{{2a - 2x}}{2}$
$\Rightarrow$ $\frac{{dy}}{{dx}} = \frac{{a - x}}{{2y}}$

$\Rightarrow$ $\frac{{dy}}{{dx}} = \frac{{2ax - 2{x^2}}}{{4xy}}$ …..(2)
From equation (1),

we get:
$2ax = 2{y^2} + {x^2}$
On substituting this value in equation (3),

we get:
$\frac{{dy}}{{dx}} = \frac{{2{y^2} + {x^2} - 2{x^2}}}{{4xy}}$
$\Rightarrow$ $\frac{{dy}}{{dx}} = \frac{{2{y^2} - {x^2}}}{{4xy}}$

Hence, the differential equation of the family of curves is given as $\frac{{dy}}{{dx}} = \frac{{2{y^2} - {x^2}}}{{4xy}}$.

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