(2 sin y + 1) dy/dx = 4 and y(0) = π/2. what is x when y = π? x =

Explanation:

Step1: Separate variables

Separate the variables in the differential - equation ((2\sin y + 1)\frac{dy}{dx}=4). We get ((2\sin y + 1)dy = 4dx).

Step2: Integrate both sides

Integrate (\int(2\sin y + 1)dy=\int4dx). The integral of (2\sin y+1) with respect to (y) is (- 2\cos y + y), and the integral of (4) with respect to (x) is (4x + C). So, (-2\cos y + y=4x + C).

Step3: Use the initial condition (y(0)=\frac{\pi}{2})

Substitute (x = 0) and (y=\frac{\pi}{2}) into (-2\cos y + y=4x + C). When (x = 0) and (y=\frac{\pi}{2}), we have (-2\cos\frac{\pi}{2}+\frac{\pi}{2}=4\times0 + C). Since (\cos\frac{\pi}{2}=0), then (C=\frac{\pi}{2}). So the equation becomes (-2\cos y + y=4x+\frac{\pi}{2}).

Step4: Find (x) when (y = \pi)

Substitute (y=\pi) into (-2\cos y + y=4x+\frac{\pi}{2}). We know that (\cos\pi=-1), so (-2\times(-1)+\pi=4x+\frac{\pi}{2}). This simplifies to (2+\pi=4x+\frac{\pi}{2}). Rearrange the equation to solve for (x): [ \begin{align*} 4x&=2+\pi-\frac{\pi}{2}\ 4x&=2+\frac{\pi}{2}\ x&=\frac{1}{2}+\frac{\pi}{8} \end{align*} ]

Answer:

(x=\frac{1}{2}+\frac{\pi}{8})