Question

The order and degree of the differential equation whose general solution is $y = c{\left( {x - c} \right)^2}$ are ?

Answer 1

In the given question, we are given a general solution of a differential equation and we are asked to compute the degree and order of the differential equation. So, we will have to form the differential equation with the help of the general solution provided to us in the question and then report the degree and order of the differential equation obtained.

The order of a differential equation is determined by the highest-order derivative; the degree is determined by the highest power on a variable.
The higher the order of the differential equation, the more arbitrary constants need to be added to the general solution.

Complete step by step solution:
The general solution of the differential equation is $y = c{\left( {x - c} \right)^2}$.
Now, to compute the order and degree of the differential equation, we will need to form the differential equation with the help of the general solution provided to us in the question.
So, we have the general solution of the differential equation as $y = c{\left( {x - c} \right)^2}$.
Differentiating both sides with respect to x, we get,
$\Rightarrow \dfrac{{dy}}{{dx}} = c\left[ {2\left( {x - c} \right)} \right] = 2c\left( {x - c} \right)$
So, we get $y' = 2c\left( {x - c} \right)$.
Now, dividing y by $y'$, we get,
$\Rightarrow \dfrac{y}{{y'}} = \dfrac{{c{{\left( {x - c} \right)}^2}}}{{2c\left( {x - c} \right)}}$
Cancelling the common factors in numerator and denominator, we get,
$\Rightarrow \dfrac{y}{{y'}} = \dfrac{{\left( {x - c} \right)}}{2}$
Shifting all the terms except c to the left side of the equation in order to isolate c and find its value, we get,
$\Rightarrow \dfrac{{2y}}{{y'}} - x = - c$
Multiplying both sides if the equation by $\left( { - 1} \right)$, we get,
$\Rightarrow c = x - \dfrac{{2y}}{{y'}}$
Hence, we can substitute the value of c in the equation $y' = 2c\left( {x - c} \right)$so as to get to the required differential equation.
So, we get,
$\Rightarrow y' = 2\left( {x - \dfrac{{2y}}{{y'}}} \right)\left( {x - \left( {x - \dfrac{{2y}}{{y'}}} \right)} \right)$
Opening up the bracket, we get,
$\Rightarrow y' = 2\left( {x - \dfrac{{2y}}{{y'}}} \right)\left( {x - x + \dfrac{{2y}}{{y'}}} \right)$
Simplifying the expression further, we get,
$\Rightarrow y' = 2\left( {x - \dfrac{{2y}}{{y'}}} \right)\left( {\dfrac{{2y}}{{y'}}} \right)$
$\Rightarrow y' = 2\left( {\dfrac{{xy' - 2y}}{{y'}}} \right)\left( {\dfrac{{2y}}{{y'}}} \right)$
$\Rightarrow {\left( {y'} \right)^3} = 4y\left( {xy' - 2y} \right)$
Hence, the order and degree of the above obtained differential equation is $1$ and $3$ respectively.

Note: Such questions require us to have a clear idea about the concepts of differential equations and differentiation. We can also notice that there is only one variable present in the given general solution of the differential equation. Hence, we can say that the degree of the differential equation would be one beforehand without finding the entire differential equation. But the differential equation must be found so as to find the degree of the differential equation.