Question

Let the solution \( y = y(x) \) of the differential equation \[\frac{dy}{dx} - y = 1 + 4 \sin x\] satisfy \( y(\pi) = 1 \). Then \( y\left( \frac{\pi}{2} \right) + 10 \) is equal to _____

Answer 1

Correct Answer: 7

Given differential equation: 
\[ \frac{dy}{dx} - y = 1 + 4 \sin x. \] 
This is a first-order linear differential equation. To solve, we use an integrating factor (IF): 
\[ \text{IF} = e^{-x}. \] 
Multiplying the entire equation by the integrating factor: \[ e^{-x} \frac{dy}{dx} - e^{-x} y = e^{-x} + 4e^{-x} \sin x. \] 
The left-hand side becomes the derivative of \(y e^{-x}\): \[ \frac{d}{dx}(y e^{-x}) = e^{-x} + 4e^{-x} \sin x. \] 
Integrating both sides: 
\[ y e^{-x} = \int \left(e^{-x} + 4e^{-x} \sin x\right) dx. \] 
Evaluating the integral: 
\[ y e^{-x} = \int e^{-x} dx + 4 \int e^{-x} \sin x \, dx. \] 

The first integral is straightforward:
\[ \int e^{-x} dx = -e^{-x}. \] 

For the second integral, we use integration by parts or known results:
\[ 4 \int e^{-x} \sin x \, dx = -2e^{-x} (\sin x + \cos x). \] 
Thus: \[ y e^{-x} = -e^{-x} - 2e^{-x} (\sin x + \cos x) + C. \] 
Multiplying through by \(e^x\): \[ y = -1 - 2(\sin x + \cos x) + C e^x. \] 
Using the initial condition \(y(\pi) = 1\): 
\[ 1 = -1 - 2(\sin \pi + \cos \pi) + C e^\pi. \]
Since \(\sin \pi = 0\) and \(\cos \pi = -1\):
\[ 1 = -1 - 2(-1) + C e^\pi \implies 1 = 1 + C e^\pi \implies C = 0. \] 
Thus, the solution simplifies to: \[ y = -1 - 2(\sin x + \cos x). \] 
Evaluating at \(x = \frac{\pi}{2}\): 
\[ y\left(\frac{\pi}{2}\right) = -1 - 2\left(\sin \frac{\pi}{2} + \cos \frac{\pi}{2}\right) = -1 - 2(1 + 0) = -3. \] 
Calculating \(y\left(\frac{\pi}{2}\right) + 10\): 
\[ y\left(\frac{\pi}{2}\right) + 10 = -3 + 10 = 7. \]