Question

The general solution of the differential equation is ${\left( {y + c} \right)^2} = cx$, where c is an arbitrary constant. What is the order and degree of the differential equation?
A) $1,2$
B) $2,2$
C) $1,1$
D) $2,1$

Answer 1

Hint: In this equation, we are given an equation ${\left( {y + c} \right)^2} = cx$ and we have been asked its order and degree. Start by differentiating the equation on both the sides. Differentiate the equation and substitute them among each other in such a way that the arbitrary constant c is eliminated. Then observe the highest order of the derivative used in the equation. Its degree will be the degree of the differential equation.

Complete step-by-step solution:
In order to tell the order and degree of the given equation, we will first differentiate both the sides and will find the equation in terms of $\dfrac{{dy}}{{dx}}$ first. Our main motive is to eliminate the arbitrary constant.
$ \Rightarrow $${\left( {y + c} \right)^2} = cx$ …. (given) (1)
Differentiating both the sides with respect to x,
$ \Rightarrow 2\left( {y + c} \right)\dfrac{{dy}}{{dx}} = c$ ….
On simplifying we will get,
$ \Rightarrow 2y\dfrac{{dy}}{{dx}} + 2c\dfrac{{dy}}{{dx}} = c$
Shifting to find the value of c,
 $ \Rightarrow 2y\dfrac{{dy}}{{dx}} = c - 2c\dfrac{{dy}}{{dx}}$
Taking c common on RHS,
$ \Rightarrow 2y\dfrac{{dy}}{{dx}} = c\left( {1 - 2\dfrac{{dy}}{{dx}}} \right)$
Finding the value of c,
$ \Rightarrow c = \dfrac{{2y\dfrac{{dy}}{{dx}}}}{{1 - 2\dfrac{{dy}}{{dx}}}}$
Putting the value of c in (1),
$ \Rightarrow $${\left( {y + \dfrac{{2y\dfrac{{dy}}{{dx}}}}{{1 - 2\dfrac{{dy}}{{dx}}}}} \right)^2} = \dfrac{{2y\dfrac{{dy}}{{dx}}}}{{1 - 2\dfrac{{dy}}{{dx}}}}x$
Taking denominator same on the LHS,
$ \Rightarrow {\left( {\dfrac{{y\left( {1 - 2\dfrac{{dy}}{{dx}}} \right)}}{{1 - 2\dfrac{{dy}}{{dx}}}} + \dfrac{{2y\dfrac{{dy}}{{dx}}}}{{1 - 2\dfrac{{dy}}{{dx}}}}} \right)^2} = \dfrac{{2y\dfrac{{dy}}{{dx}}}}{{1 - 2\dfrac{{dy}}{{dx}}}}x$
Simplifying,
$ \Rightarrow {\left( {\dfrac{{y - 2y\dfrac{{dy}}{{dx}} + 2y\dfrac{{dy}}{{dx}}}}{{1 - 2\dfrac{{dy}}{{dx}}}}} \right)^2} = \dfrac{{2y\dfrac{{dy}}{{dx}}}}{{1 - 2\dfrac{{dy}}{{dx}}}}x$
$ \Rightarrow {\left( y \right)^2} = \dfrac{{2y\dfrac{{dy}}{{dx}}}}{{1 - 2\dfrac{{dy}}{{dx}}}}x \times {\left( {1 - 2\dfrac{{dy}}{{dx}}} \right)^2}$
$ \Rightarrow {y^2} = 2xy\dfrac{{dy}}{{dx}} \times \left( {1 - 2\dfrac{{dy}}{{dx}}} \right)$
On multiplying in RHS,
$ \Rightarrow {y^2} = 2xy\dfrac{{dy}}{{dx}} - 4xy{\left( {\dfrac{{dy}}{{dx}}} \right)^2}$
Now, we have our equation in the simplest form. So, we can now tell the order and degree of our differential equation.
Order of D.E = It is the order of the highest derivative used in the equation.
In this equation, order of D.E = 1
Degree of D.E = It is the degree of the highest order derivative used in the question.
In this equation, degree of D.E = 2

$\therefore $ Our answer is option (A) 1,2

Note: One should remember that, in mathematics, a differential equation is an equation that relates one or more functions and their derivatives. In applications, the functions generally represent physical equations that define a relationship between the two. Such relations are common, therefore differential equations play a prominent role in many disciplines including engineering, physics, economics, and biology. The order of the derivative can be founded by without actually solving the question. The order of the equation is the number of constants in the equation. Here, in the equation, we have only 1 constant so our order is 1.