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Question: In general solutions of a differential equation is \({(y + c)^2} = cx,\) where c is an arbitrary con...

In general solutions of a differential equation is (y+c)2=cx,{(y + c)^2} = cx, where c is an arbitrary constant, then the order and degree of the differential equation.
A) 1, 2
B) 2, 1
C) 1, 1
D) None of these

Explanation

Solution

According to given in the question we have to determine the order and degree of the differential equation (y+c)2=cx,{(y + c)^2} = cx, so, first of all we have to understand about order and degree as explained below:
Order: Order of a differential equation is the order of the highest order derivatives present in the equation or given in the equation.
Degree: The degree of the differential equation is the power of its highest derivatives after we have obtained the rational and integral in all of its derivatives.
Now, we have to find the differentiation with respect to x on the both sides of the differential equation. After that we have to substitute the c in the general solution. Hence, after that we can obtain the order and degree as asked in the solution.

Formula used: ⇒dxndx=nxn−1................(A) ⇒dxdx=1................(B) \Rightarrow \dfrac{{d{x^n}}}{{dx}} = n{x^{n - 1}}................(A) \\\ \Rightarrow \dfrac{{dx}}{{dx}} = 1................(B)

Complete step-by-step answer:
Step 1: First of all we have to find the differentiation with respect to x on the both sides of the differential equation as mentioned in the solution hint. Hence,
⇒d(y+c)2dx=dcxdx\Rightarrow \dfrac{{d{{(y + c)}^2}}}{{dx}} = \dfrac{{dcx}}{{dx}}
Step 2: Now, to solve the differential equation as obtained in the solution step 1 we have to use the formula (A), and formula (B) as mentioned in the solution hint. Hence,
⇒2(y+c)dydx=c\Rightarrow 2(y + c)\dfrac{{dy}}{{dx}} = c
Step 3: Now, we have to let that dydx=Y\dfrac{{dy}}{{dx}} = Y. Hence, on substituting the values in the expression as obtained in the solution step 2.
⇒c=2yY1−2Y\Rightarrow c = \dfrac{{2yY}}{{1 - 2Y}}
Step 4: Now, on substituting the value of c as obtained in the step 3 in the expression as given in the question.
⇒(y+2yY1−2Y)2=2yY1−2Yx\Rightarrow {\left( {y + \dfrac{{2yY}}{{1 - 2Y}}} \right)^2} = \dfrac{{2yY}}{{1 - 2Y}}x
Step 5: Now, to solve the expression as obtained in the step 4 we have to apply the formula (C) as mentioned in the solution hint,
⇒y2=2yYx(1−2Y)\Rightarrow {y^2} = 2yYx(1 - 2Y)
On multiplying each terms of the obtained expression,
=2yxY−4yxY2= 2yxY - 4yx{Y^2}
Step 6: Now, as explained about the order and degree in the solution hint we can determine the order and degree for the expression as obtained in the step 5. Hence,
Order = 1
Degree = 2
Final solution: Hence, with the help of formulas as mentioned in the solution hint we have obtained the order and degree for the differential equation which are 1, and 2.

Therefore the correct option is (A).

Note: The order of the differential equation depends on the derivative of the highest order in the equation and the degree of the differential equation similarly can be determined by the highest exponent on any variable in the differential equation.
All the derivatives in the differential equation are free from fractional powers which can be positive as well as negative if any.