Question

The differential equation representing the family of curves ${y^2} = 2c\left( {x + \sqrt c } \right)$, where $c$ is a positive perimeter, is of
A. Order 1, degree 3
B. Order 2, degree 3
C. Degree 3, order 2
D. None of these

Answer 1

Here we will proceed by differentiating on both sides with respective to $x$ and from this we evaluate the value of the constant $c$. Further substitute this value in the given family of curves to get the required solution of the given problem.

Complete step-by-step answer:
Given family of curves is ${y^2} = 2c\left( {x + \sqrt c } \right)..............................................................\left( 1 \right)$
Differentiating ${y^2} = 2c\left( {x + \sqrt c } \right)$ on both sides w.r.t $x$ we have

$$\Rightarrow \dfrac{d}{{dx}}\left( {{y^2}} \right) = \dfrac{d}{{dx}}\left( {2c\left( {x + \sqrt c } \right)} \right) \\\ \Rightarrow 2y\dfrac{{dy}}{{dx}} = 2c\dfrac{d}{{dx}}\left( {x + \sqrt c } \right) \\\ \Rightarrow 2y\dfrac{{dy}}{{dx}} = 2c\left[ {\dfrac{d}{{dx}}\left( x \right) + \dfrac{d}{{dx}}\left( {\sqrt c } \right)} \right] \\\ \Rightarrow 2y\dfrac{{dy}}{{dx}} = 2c\left[ {1 + 0} \right] \\\ \Rightarrow 2y\dfrac{{dy}}{{dx}} = 2c \\\ \therefore c = y\dfrac{{dy}}{{dx}}.....................................................................\left( 2 \right) \\\$$

Substituting equation (2) in equation (1), we have

$$\Rightarrow {y^2} = 2\left( {y\dfrac{{dy}}{{dx}}} \right)\left( {x + \sqrt {y\dfrac{{dy}}{{dx}}} } \right) \\\ \Rightarrow {y^2} = 2xy\dfrac{{dy}}{{dx}} + 2x\dfrac{{dy}}{{dx}}\sqrt {y\dfrac{{dy}}{{dx}}} \\\ \Rightarrow {y^2} = 2xy\dfrac{{dy}}{{dx}} + 2x\sqrt y {\left( {\dfrac{{dy}}{{dx}}} \right)^{\dfrac{3}{2}}} \\\ \Rightarrow {y^2} - 2xy\dfrac{{dy}}{{dx}} = 2x\sqrt y {\left( {\dfrac{{dy}}{{dx}}} \right)^{\dfrac{3}{2}}} \\\$$

Squaring on both sides, we get

$$\Rightarrow {\left( {{y^2} - 2xy\dfrac{{dy}}{{dx}}} \right)^2} = {\left( {2x\sqrt y {{\left( {\dfrac{{dy}}{{dx}}} \right)}^{\dfrac{3}{2}}}} \right)^2} \\\ \Rightarrow {y^4} - 4x{y^3}\dfrac{{dy}}{{dx}} + 4{x^2}{y^2}{\left( {\dfrac{{dy}}{{dx}}} \right)^2} = 4{x^2}y{\left( {\dfrac{{dy}}{{dx}}} \right)^3} \\\ \Rightarrow 4{x^2}y{\left( {\dfrac{{dy}}{{dx}}} \right)^3} - 4{x^2}{y^2}{\left( {\dfrac{{dy}}{{dx}}} \right)^2} + 4x{y^3}\dfrac{{dy}}{{dx}} - {y^4} = 0 \\\$$

Hence the solution of the given family of curves is $4{x^2}y{\left( {\dfrac{{dy}}{{dx}}} \right)^3} - 4{x^2}{y^2}{\left( {\dfrac{{dy}}{{dx}}} \right)^2} + 4x{y^3}\dfrac{{dy}}{{dx}} - {y^4} = 0$
From the above differential equation clearly, the order is 1 and degree is 3.
Thus, the correct option is A. Order 1, degree 3

So, the correct answer is “Option A”.

Note: The degree of the differential equation is represented by the power of the highest order derivative of the given differential equation. Order of a differential equation is the highest derivative present in the given differential equation.