Question
The degree of the differential equation ${\left[ {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} \right]^{3/2}} = \frac{{{d^2}y}}{{d{x^2}}}$ is
- (a) 4
- (b) $\frac{3}{2}$
- (c) not defined
- (d) 2 ✓ Correct
Differential Equations — Class 12 Maths Solution
Step-by-step Solution
Given that ${\left[ {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} \right]^{3/2}} = \frac{{{d^2}y}}{{d{x^2}}}$
On squaring both sides,
we get
${\left[ {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} \right]^3} = {\left( {\frac{{{d^2}y}}{{d{x^2}}}} \right)^2}$
So, the degree of differential equation is 2 .
NCERT & Exemplar solution for CBSE Class 12 Mathematics, Differential Equations. Curated by Sachin Sharma. Free for all students.