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

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 $y\left(\frac{\pi}{2}\right) + 10$:
$y\left(\frac{\pi}{2}\right) + 10 = -3 + 10 = 7.$