Question

The order and the degree of the differential equation $2\frac{dy}{dx}-3x=\left(2y-x\frac{dy}{dx}\right)^{-3}$ respectively, are

• A. 1 and 1
• B. 1 and 3
• C. 1 and 4
• D. 2 and 3
• E. 2 and 4

Hint:

Logic Tip: You rarely need to expand the entire binomial. Simply take the highest derivative term inside the parentheses, raise it to the exponent outside, and multiply it by any other highest derivative terms outside the parentheses to find the final degree.

Answer 1

The Correct Option is C

Concept:
The order of a differential equation is the order of the highest derivative present. The degree of a differential equation is the highest power (exponent) of the highest order derivative, but \textit{only after} the equation has been cleared of fractional or negative exponents involving derivatives.
Step 1: Eliminate the negative exponent.
The given equation is: $$2\frac{dy}{dx} - 3x = \frac{1}{\left(2y - x\frac{dy}{dx}\right)^3}$$ To make it a polynomial equation in its derivatives, cross-multiply by the denominator: $$\left(2y - x\frac{dy}{dx}\right)^3 \left(2\frac{dy}{dx} - 3x\right) = 1$$
Step 2: Determine the order.
The only derivative present in the entire equation is $\frac{dy}{dx}$, which is a first derivative. Therefore, the order is 1.
Step 3: Determine the degree.
We need to find the highest power of $\frac{dy}{dx}$ in the expanded polynomial. Consider the highest power term from the cubic expansion: $$\left(2y - x\frac{dy}{dx}\right)^3 = (2y)^3 - 3(2y)^2\left(x\frac{dy}{dx}\right) + 3(2y)\left(x\frac{dy}{dx}\right)^2 - \left(x\frac{dy}{dx}\right)^3$$ The term with the highest power of the derivative here is $-x^3\left(\frac{dy}{dx}\right)^3$. Now multiply this leading term by the term containing the derivative in the second bracket, $2\frac{dy}{dx}$: $$\left[ -x^3\left(\frac{dy}{dx}\right)^3 \right] \cdot \left[ 2\frac{dy}{dx} \right] = -2x^3\left(\frac{dy}{dx}\right)^4$$ The highest power to which the derivative is raised in the fully expanded equation is 4. Therefore, the degree is 4.