The order and degree of the differential equation y d x 2 + ( dx ) 1/4 + x 1/5 = 0 respectively, are

Given that, y d x 2 + ( dx ) 1/4 = - x 1/5 y d x 2 + ( dx ) 1/4 = - x 1/5 dx ) 1/4 = - ( x 1/5 + y d x 2 ) On squaring both sides, we get ( dx ) 1/2 = ( x 1/5 + y d x 2 ) 2 Again, on squaring both sides, we have dx = ( x 1/5 + y d x 2 ) 4 order = 2 , degree = 4

Differential Equations — Class 12 Maths Solution

exemplar objective MCQ NCERT EXEMP.Q.36,Page.195

Question

The order and degree of the differential equation$\frac{{{d^2}y}}{{d{x^2}}} + {\left( {\frac{{dy}}{{dx}}} \right)^{1/4}} + {x^{1/5}} = 0$
respectively, are

(a) 2 and 4 ✓ Correct
(b) 2 and 2
(c) 2 and 3
(d) 3 and 3

Step-by-step Solution

Given that, $\frac{{{d^2}y}}{{d{x^2}}} + {\left( {\frac{{dy}}{{dx}}} \right)^{1/4}} = - {x^{1/5}}$

$\Rightarrow$ $\frac{{{d^2}y}}{{d{x^2}}} + {\left( {\frac{{dy}}{{dx}}} \right)^{1/4}} = - {x^{1/5}}$

$\Rightarrow$ ${\left( {\frac{{dy}}{{dx}}} \right)^{1/4}} = - \left( {{x^{1/5}} + \frac{{{d^2}y}}{{d{x^2}}}} \right)$

On squaring both sides,

we get
${\left( {\frac{{dy}}{{dx}}} \right)^{1/2}} = {\left( {{x^{1/5}} + \frac{{{d^2}y}}{{d{x^2}}}} \right)^2}$

Again, on squaring both sides,

we have
$\frac{{dy}}{{dx}} = {\left( {{x^{1/5}} + \frac{{{d^2}y}}{{d{x^2}}}} \right)^4}$

order $= 2$, degree $= 4$

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