Q114 / 180 · COMEDK 2023 (Evening Shift) · Mathematics
The sum of the degree and order of the following differential equation $\left[1-\left(\dfrac{d y}{d x}\right)^2\right]^{\dfrac{3}{2}}=k x \dfrac{d^2 y}{d x^2}$
- A3
- Correct answerB4
- C$\dfrac{3}{2}$
- D$\dfrac{5}{2}$
Solution
The given differential equation is $\left[1 - \left(\dfrac{dy}{dx}\right)^2\right]^{\dfrac{3}{2}} = kx \dfrac{d^2y}{dx^2}$.
To determine the order and degree, the equation must be expressed as a polynomial in terms of the derivatives. Squaring both sides to eliminate the fractional exponent, we get:
$\left[1 - \left(\dfrac{dy}{dx}\right)^2\right]^3 = (kx)^2 \left(\dfrac{d^2y}{dx^2}\right)^2$.
The order of the differential equation is the highest derivative present, which is $\dfrac{d^2y}{dx^2}$. Thus, the order is 2.
The degree of the differential equation is the power of the highest order derivative when the equation is written as a polynomial in derivatives. Here, the power of $\dfrac{d^2y}{dx^2}$ is 2. Thus, the degree is 2.
The sum of the degree and the order is $2 + 2 = 4$.
Answer: 4
To determine the order and degree, the equation must be expressed as a polynomial in terms of the derivatives. Squaring both sides to eliminate the fractional exponent, we get:
$\left[1 - \left(\dfrac{dy}{dx}\right)^2\right]^3 = (kx)^2 \left(\dfrac{d^2y}{dx^2}\right)^2$.
The order of the differential equation is the highest derivative present, which is $\dfrac{d^2y}{dx^2}$. Thus, the order is 2.
The degree of the differential equation is the power of the highest order derivative when the equation is written as a polynomial in derivatives. Here, the power of $\dfrac{d^2y}{dx^2}$ is 2. Thus, the degree is 2.
The sum of the degree and the order is $2 + 2 = 4$.
Answer: 4
Hence, the correct answer is option: 'B'.