question solve the following initial - value problem. f(x)=cos(x)-sin(x), f(π/3)=√3/2 + 7/2 provide your…

Explanation:

Step1: Integrate $f'(x)$

Integrate $\cos(x)-\sin(x)$ with respect to $x$. We know that $\int\cos(x)dx=\sin(x)+C_1$ and $\int\sin(x)dx = -\cos(x)+C_2$. So, $f(x)=\int(\cos(x)-\sin(x))dx=\sin(x)+\cos(x)+C$.

Step2: Use the initial - condition

Given $f(\frac{\pi}{3})=\frac{\sqrt{3}}{2}+\frac{7}{2}$. Substitute $x = \frac{\pi}{3}$ into $f(x)=\sin(x)+\cos(x)+C$. We have $\sin(\frac{\pi}{3})+\cos(\frac{\pi}{3})+C=\frac{\sqrt{3}}{2}+\frac{7}{2}$. Since $\sin(\frac{\pi}{3})=\frac{\sqrt{3}}{2}$ and $\cos(\frac{\pi}{3})=\frac{1}{2}$, then $\frac{\sqrt{3}}{2}+\frac{1}{2}+C=\frac{\sqrt{3}}{2}+\frac{7}{2}$.

Step3: Solve for $C$

Subtract $\frac{\sqrt{3}}{2}+\frac{1}{2}$ from both sides of the equation $\frac{\sqrt{3}}{2}+\frac{1}{2}+C=\frac{\sqrt{3}}{2}+\frac{7}{2}$. We get $C = 3$.

Answer:

$\sin(x)+\cos(x)+3$