\( 2, 3 \)
Step 1: Identifying the Order of the Differential Equation
The order of a differential equation is the highest order derivative present in the equation. \[ \frac{dy}{dx} + \left( \frac{d^2y}{dx^2} + 2 \right)^{\frac{1}{2}} + \frac{d^3y}{dx^3} + 5 = 0. \] Here, the highest order derivative is \( \frac{d^3y}{dx^3} \), which means the order of the differential equation is: \[ \mathbf{3}. \]
Step 2: Identifying the Degree of the Differential Equation
The degree of a differential equation is defined as the highest exponent of the highest order derivative after removing radicals and fractions. In this equation, the term \( \left( \frac{d^2y}{dx^2} + 2 \right)^{\frac{1}{2}} \) contains a fractional power. To determine the degree, we must first eliminate this radical by squaring both sides. After squaring, the highest exponent of \( \frac{d^3y}{dx^3} \) (the highest order derivative) is found to be: \[ \mathbf{2}. \]
Step 3: Conclusion
Thus, the correct answer is: \[ \mathbf{2, 2}. \]
For the beam and loading shown in the figure, the second derivative of the deflection curve of the beam at the mid-point of AC is given by \( \frac{\alpha M_0}{8EI} \). The value of \( \alpha \) is ........ (rounded off to the nearest integer).
If the function \[ f(x) = \begin{cases} \frac{2}{x} \left( \sin(k_1 + 1)x + \sin(k_2 -1)x \right), & x<0 \\ 4, & x = 0 \\ \frac{2}{x} \log_e \left( \frac{2 + k_1 x}{2 + k_2 x} \right), & x>0 \end{cases} \] is continuous at \( x = 0 \), then \( k_1^2 + k_2^2 \) is equal to:
The integral is given by:
\[ 80 \int_{0}^{\frac{\pi}{4}} \frac{\sin\theta + \cos\theta}{9 + 16 \sin 2\theta} d\theta \]
is equals to?