Question

If the order and degree of the differential equation \(x \frac{d^2 y}{dx^2} = \left(1 + \left(\frac{d^2 y}{dx^2}\right)^2\right)^{-1/2}\) are \(k\) and \(l\) respectively, then \(k, l\) are the roots of

• A. \(x^2 - 5x + 6 = 0 \)
• B. \(x^2 - 3x + 2 = 0 \)
• C. \(x^2 - 7x + 12 = 0 \)
• D. \(x^2 - 6x + 8 = 0 \)

Hint:

To find the order and degree of a differential equation: 1. Clear all radicals and fractions involving the derivatives. This usually involves raising both sides to a suitable power and/or multiplying by denominators. 2. Order: Identify the highest order derivative present in the equation. This is the order of the differential equation. 3. Degree: Once the equation is in polynomial form with respect to derivatives, the degree is the highest power of the highest order derivative.

Answer 1

The Correct Option is D

Step 1: Rewrite the given differential equation to eliminate radicals and fractions.
The given differential equation is: \[ x \frac{d^2 y}{dx^2} = \left(1 + \left(\frac{d^2 y}{dx^2}\right)^2\right)^{-1/2} \] The term with the negative fractional exponent can be rewritten as a reciprocal with a positive fractional exponent: \[ x \frac{d^2 y}{dx^2} = \frac{1}{\left(1 + \left(\frac{d^2 y}{dx^2}\right)^2\right)^{1/2}} \] To remove the square root (which is represented by the power of 1/2), square both sides of the equation: \[ \left(x \frac{d^2 y}{dx^2}\right)^2 = \left(\frac{1}{\left(1 + \left(\frac{d^2 y}{dx^2}\right)^2\right)^{1/2}}\right)^2 \] \[ x^2 \left(\frac{d^2 y}{dx^2}\right)^2 = \frac{1}{1 + \left(\frac{d^2 y}{dx^2}\right)^2} \] Now, multiply both sides by the denominator \(\left(1 + \left(\frac{d^2 y}{dx^2}\right)^2\right)\) to clear the fraction: \[ x^2 \left(\frac{d^2 y}{dx^2}\right)^2 \left(1 + \left(\frac{d^2 y}{dx^2}\right)^2\right) = 1 \] Distribute \(x^2 \left(\frac{d^2 y}{dx^2}\right)^2\): \[ x^2 \left(\frac{d^2 y}{dx^2}\right)^2 + x^2 \left(\frac{d^2 y}{dx^2}\right)^4 = 1 \] Rearrange the terms to form a polynomial in terms of the derivatives: \[ x^2 \left(\frac{d^2 y}{dx^2}\right)^4 + x^2 \left(\frac{d^2 y}{dx^2}\right)^2 - 1 = 0 \] Step 2: Determine the order and degree of the differential equation.
Order (k): The order of a differential equation is the order of the highest derivative present in the equation. In the given equation, the highest derivative is \(\frac{d^2 y}{dx^2}\), which is a second-order derivative. So, \(k = 2\). Degree (l): The degree of a differential equation is the power of the highest order derivative when the differential equation is free from radicals and fractions, and derivatives are present in polynomial form. In our simplified equation, the highest order derivative is \(\frac{d^2 y}{dx^2}\), and its highest power is 4 (from the term \(x^2 \left(\frac{d^2 y}{dx^2}\right)^4\)). So, \(l = 4\). 
Step 3: Form the quadratic equation whose roots are \(k\) and \(l\).
The roots of the quadratic equation are \(k=2\) and \(l=4\).
A quadratic equation with roots \(\alpha\) and \(\beta\) is given by \(x^2 - (\alpha + \beta)x + \alpha\beta = 0\).
Sum of the roots: \(k + l = 2 + 4 = 6\).
Product of the roots: \(k \cdot l = 2 \cdot 4 = 8\). Substitute these values into the quadratic equation formula: \[ x^2 - (6)x + 8 = 0 \] \[ x^2 - 6x + 8 = 0 \] This equation matches option (4). The final answer is $\boxed{x^2 - 6x + 8 = 0}$.