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

Answer 2

Step 1: Write the given differential equation.
The given equation is: \[ \frac{dy}{dx} - y = 1 + 4\sin x \] This is a linear differential equation of the form: \[ \frac{dy}{dx} + P(x)y = Q(x) \] where \( P(x) = -1 \) and \( Q(x) = 1 + 4\sin x. \)

Step 2: Find the integrating factor (I.F).
The integrating factor is: \[ I.F = e^{\int P(x)dx} = e^{\int (-1)dx} = e^{-x}. \]

Step 3: Multiply the equation by the integrating factor.
\[ e^{-x}\frac{dy}{dx} - e^{-x}y = (1 + 4\sin x)e^{-x}. \] The left-hand side can be written as the derivative of \( y \cdot e^{-x} \): \[ \frac{d}{dx}(y e^{-x}) = (1 + 4\sin x)e^{-x}. \]

Step 4: Integrate both sides.
\[ y e^{-x} = \int (1 + 4\sin x)e^{-x}dx + C. \] We separate the terms: \[ y e^{-x} = \int e^{-x}dx + 4\int e^{-x}\sin x \, dx + C. \]

Step 5: Solve each integral.
First integral: \[ \int e^{-x}dx = -e^{-x}. \] Second integral: \[ \int e^{-x}\sin x \, dx = \frac{e^{-x}}{2}(-\sin x - \cos x). \] (Using the standard result \(\int e^{ax}\sin bx \, dx = \frac{e^{ax}(a\sin bx - b\cos bx)}{a^2 + b^2}\), with \(a=-1, b=1\).)
So: \[ \int e^{-x}\sin x \, dx = \frac{e^{-x}(-\sin x - \cos x)}{2}. \] Therefore: \[ 4\int e^{-x}\sin x \, dx = 2e^{-x}(-\sin x - \cos x) = -2e^{-x}(\sin x + \cos x). \]

Step 6: Substitute back.
\[ y e^{-x} = -e^{-x} - 2e^{-x}(\sin x + \cos x) + C. \] Multiply through by \( e^x \): \[ y = -1 - 2(\sin x + \cos x) + Ce^x. \]

Step 7: Apply the initial condition \( y(\pi) = 1 \).
\[ 1 = -1 - 2(\sin \pi + \cos \pi) + Ce^{\pi}. \] \[ 1 = -1 - 2(0 - 1) + Ce^{\pi}. \] \[ 1 = -1 + 2 + Ce^{\pi} \Rightarrow Ce^{\pi} = 0 \Rightarrow C = 0. \] Thus, \[ y = -1 - 2(\sin x + \cos x). \]

Step 8: Find \( y\left(\frac{\pi}{2}\right) + 10 \).
\[ y\left(\frac{\pi}{2}\right) = -1 - 2(\sin \frac{\pi}{2} + \cos \frac{\pi}{2}) = -1 - 2(1 + 0) = -3. \] \[ y\left(\frac{\pi}{2}\right) + 10 = -3 + 10 = 7. \]

Final Answer:
\[ \boxed{7} \]