To find the degree of a differential equation, first make sure all derivatives are polynomial, and eliminate any fractional powers or roots. The degree is determined by the highest power of the highest-order derivative in the equation.
The correct answer is: (B): 6.
We are tasked with determining the degree of the differential equation:
\(1 + \left(\frac{dy}{dx}\right)^2 + \left(\frac{d^2y}{dx^2}\right)^2 = \sqrt[3]{\frac{d^2y}{dx^2} + 1}\)
Step 1: Express the equation in terms of derivatives
To find the degree, we need to ensure that the equation is polynomial in the derivatives of \( y \). Start by clearing the cube root by raising both sides of the equation to the power of 3:
\( \left( 1 + \left(\frac{dy}{dx}\right)^2 + \left(\frac{d^2y}{dx^2}\right)^2 \right)^3 = \frac{d^2y}{dx^2} + 1 \)
Step 2: Eliminate the cube root and simplify
By simplifying both sides of the equation, we can rewrite the equation in a form where the derivatives are only raised to integer powers. This step essentially removes the fractional powers and simplifies the equation to a polynomial in \( \frac{dy}{dx} \) and \( \frac{d^2y}{dx^2} \). The left side of the equation becomes a polynomial, while the right side has only the second derivative of \( y \), ensuring that we have a polynomial equation in terms of the derivatives.
Step 3: Identify the degree of the equation
The degree of a differential equation is defined as the highest power of the highest order derivative in the equation. In this case, after simplifying the equation, the highest order derivative is \( \frac{d^2y}{dx^2} \), and the highest power of this derivative is 6, making the degree of the equation 6.
Conclusion:
The degree of the differential equation is (B): 6.
A wooden block of mass M lies on a rough floor. Another wooden block of the same mass is hanging from the point O through strings as shown in the figure. To achieve equilibrium, the coefficient of static friction between the block on the floor and the floor itself is
In an experiment to determine the figure of merit of a galvanometer by half deflection method, a student constructed the following circuit. He applied a resistance of \( 520 \, \Omega \) in \( R \). When \( K_1 \) is closed and \( K_2 \) is open, the deflection observed in the galvanometer is 20 div. When \( K_1 \) is also closed and a resistance of \( 90 \, \Omega \) is removed in \( S \), the deflection becomes 13 div. The resistance of galvanometer is nearly: