If \( y = \sin x + A \cos x \) is the general solution of \[ \frac{dy}{dx} + f(x)y = \sec x, \] then an integrating factor of the differential equation is:
\( \sin x \)
Step 1: Identifying the Given Differential Equation
The given general solution is: \[ y = \sin x + A \cos x. \] Differentiating both sides: \[ \frac{dy}{dx} = \cos x - A \sin x. \] From the given differential equation: \[ \frac{dy}{dx} + f(x)y = \sec x. \] Substituting \( y = \sin x + A \cos x \): \[ \cos x - A \sin x + f(x) (\sin x + A \cos x) = \sec x. \] Rearrange: \[ \cos x - A \sin x + f(x) \sin x + A f(x) \cos x = \sec x. \]
Step 2: Finding Integrating Factor (IF)
A standard linear differential equation is of the form: \[ \frac{dy}{dx} + P(x)y = Q(x). \] The integrating factor (IF) is given by: \[ e^{\int P(x) dx}. \] From the given form, we identify: \[ P(x) = \tan x. \] Thus, the integrating factor is: \[ e^{\int \tan x dx} = e^{\ln |\sec x|} = \sec x. \]
Step 3: Conclusion
Thus, the integrating factor is: \[ \boxed{\sec x}. \]
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?