(d) use separation of variables to find y = f(x), the particular solution to the differential equation dy/dx…

(d) use separation of variables to find y = f(x), the particular solution to the differential equation dy/dx = 1/2 sin(π/2 x)√(y + 7) with the initial condition f(1) = 2. 5 points 3. the graph of the differentiable function f, shown for -6 ≤ x ≤ 7, has a horizontal tangent at x = -2 and is linear for 0 ≤ x ≤ 7. let r be the region in the second quadrant bounded by the graph of f, the vertical line x = -6, and the x - and y - axes. region r has area 12. (a) the function g is defined by g(x)=∫₀ˣf(t)dt. find the values of g(-6), g(4), and g(6).

(d) use separation of variables to find y = f(x), the particular solution to the differential equation dy/dx = 1/2 sin(π/2 x)√(y + 7) with the initial condition f(1) = 2. 5 points 3. the graph of the differentiable function f, shown for -6 ≤ x ≤ 7, has a horizontal tangent at x = -2 and is linear for 0 ≤ x ≤ 7. let r be the region in the second quadrant bounded by the graph of f, the vertical line x = -6, and the x - and y - axes. region r has area 12. (a) the function g is defined by g(x)=∫₀ˣf(t)dt. find the values of g(-6), g(4), and g(6).

Answer

Explanation:

Step1: Separate variables

Separate the variables in the differential equation $\frac{dy}{dx}=\frac{1}{2}\sin(\frac{\pi}{2}x)\sqrt{y + 7}$. We get $\frac{dy}{\sqrt{y+7}}=\frac{1}{2}\sin(\frac{\pi}{2}x)dx$.

Step2: Integrate both sides

Integrate the left - hand side $\int\frac{dy}{\sqrt{y + 7}}$ and the right - hand side $\int\frac{1}{2}\sin(\frac{\pi}{2}x)dx$. For $\int\frac{dy}{\sqrt{y + 7}}$, let $u=y + 7$, $du=dy$, then $\int\frac{dy}{\sqrt{y + 7}}=\int u^{-\frac{1}{2}}du=2\sqrt{u}=2\sqrt{y + 7}+C_1$. For $\int\frac{1}{2}\sin(\frac{\pi}{2}x)dx$, let $t=\frac{\pi}{2}x$, $dt=\frac{\pi}{2}dx$, then $\int\frac{1}{2}\sin(\frac{\pi}{2}x)dx=\frac{1}{\pi}\int\sin(t)dt=-\frac{1}{\pi}\cos(t)+C_2=-\frac{1}{\pi}\cos(\frac{\pi}{2}x)+C$. So, $2\sqrt{y + 7}=-\frac{1}{\pi}\cos(\frac{\pi}{2}x)+C$.

Step3: Use the initial condition

Given $f(1) = 2$, substitute $x = 1$ and $y=2$ into $2\sqrt{y + 7}=-\frac{1}{\pi}\cos(\frac{\pi}{2}x)+C$. $2\sqrt{2 + 7}=-\frac{1}{\pi}\cos(\frac{\pi}{2}\times1)+C$. $2\times3=0 + C$, so $C = 6$.

Step4: Solve for y

We have $2\sqrt{y + 7}=-\frac{1}{\pi}\cos(\frac{\pi}{2}x)+6$. $\sqrt{y + 7}=\frac{1}{2}(6-\frac{1}{\pi}\cos(\frac{\pi}{2}x))$. Square both sides: $y=\left(\frac{6-\frac{1}{\pi}\cos(\frac{\pi}{2}x)}{2}\right)^2-7$.

Answer:

$y=\left(\frac{6-\frac{1}{\pi}\cos(\frac{\pi}{2}x)}{2}\right)^2-7$