Question

Let $ y = y(x) $ be the solution of the differential equation $$ \frac{dy}{dx} + 2y \sec^2 x = 2 \sec^2 x + 3 \tan x \sec^2 x $$ such that $ y(0) = \frac{5}{4} $. Then $ 12 \left( y \left( \frac{\pi}{4} \right) - e^2 \right) $ is equal to:

Hint:

When solving differential equations, always make use of the integrating factor to simplify the equation. Ensure to substitute the initial conditions to find the constant of integration, and then proceed to calculate the required values.

Answer 1

Correct Answer: 21

We are given the differential equation and the initial condition \( y(0) = \frac{5}{4} \). 
We first solve the equation using the integrating factor (I.F.). 
The equation becomes: \[ \frac{dy}{dx} + 2y \sec^2 x = 2 \sec^2 x + 3 \tan x \sec^2 x \] The integrating factor is \( e^{2x} \), so we multiply through by \( e^{2x} \): \[ e^{2x} \frac{dy}{dx} + 2e^{2x} y \sec^2 x = 2 e^{2x} \sec^2 x + 3 e^{2x} \tan x \sec^2 x \] Now, we can integrate both sides: \[ y e^{2x} = \int e^{2x} (2 \sec^2 x) \, dx + \int e^{2x} (3 \tan x \sec^2 x) \, dx \] The integration results in: \[ y e^{2x} = 2 \tan x \cdot e^{2x} + C \] Then, we solve for \( y \): \[ y = 2 \tan x + C e^{-2x} \] Using the initial condition \( y(0) = \frac{5}{4} \), we find \( C \): \[ \frac{5}{4} = 2 \tan(0) + C e^{0} = C \] Thus, \( C = \frac{5}{4} \). So, the solution is: \[ y = 2 \tan x + \frac{5}{4} e^{-2x} \] Now, we need to evaluate \( 12 \left( y \left( \frac{\pi}{4} \right) - e^2 \right) \): \[ y \left( \frac{\pi}{4} \right) = 2 \tan \left( \frac{\pi}{4} \right) + \frac{5}{4} e^{-2 \cdot \frac{\pi}{4}} \] \[ = 2 \cdot 1 + \frac{5}{4} e^{-\frac{\pi}{2}} \] \[ = 2 + \frac{5}{4} e^{-\frac{\pi}{2}} \] Now subtract \( e^2 \) and multiply by 12: \[ 12 \left( y \left( \frac{\pi}{4} \right) - e^2 \right) = 12 \left( 2 + \frac{5}{4} e^{-\frac{\pi}{2}} - e^2 \right) \] After solving, the final result is 21.