find the function f that satisfies the following differential equations and initial conditions. f″(x)=cos x…

Explanation:

Step1: Integrate $F''(x)$ to get $F'(x)$

Integrate $\cos x$ with respect to $x$. We know that $\int\cos xdx=\sin x + C_1$. So $F'(x)=\sin x + C_1$.

Step2: Use the initial - condition $F'(\pi)=4$ to find $C_1$

Substitute $x = \pi$ into $F'(x)$: $F'(\pi)=\sin\pi + C_1$. Since $\sin\pi=0$, and $F'(\pi) = 4$, we have $0 + C_1=4$, so $C_1 = 4$. Then $F'(x)=\sin x+4$.

Step3: Integrate $F'(x)$ to get $F(x)$

Integrate $\sin x + 4$ with respect to $x$. We know that $\int(\sin x + 4)dx=-\cos x+4x + C_2$. So $F(x)=-\cos x+4x + C_2$.

Step4: Use the initial - condition $F(\pi)=3$ to find $C_2$

Substitute $x = \pi$ into $F(x)$: $F(\pi)=-\cos\pi+4\pi + C_2$. Since $\cos\pi=-1$, we have $-(-1)+4\pi + C_2=3$. That is $1 + 4\pi+C_2=3$, so $C_2=2 - 4\pi$.

Step5: Write the final function $F(x)$

$F(x)=-\cos x+4x+2 - 4\pi$.

Answer:

$-\cos x+4x + 2-4\pi$