Question

1. $f(t)=2e^{t}+3sin t$, $f(0)=0$, $f(pi)=0$

Explanation:

Step1: Integrate $f''(t)$ to get $f'(t)$

\[f'(t)=\int(2e^{t}+3\sin t)dt = 2e^{t}-3\cos t + C_1\]

Step2: Integrate $f'(t)$ to get $f(t)$

\[f(t)=\int(2e^{t}-3\cos t + C_1)dt=2e^{t}-3\sin t + C_1t + C_2\]

Step3: Use the condition $f(0) = 0$

\[f(0)=2e^{0}-3\sin(0)+C_1\times0 + C_2=2 + C_2=0\]
So, $C_2=-2$.

Step4: Use the condition $f(\pi)=0$ with $C_2 = - 2$

\[f(\pi)=2e^{\pi}-3\sin(\pi)+C_1\pi-2=2e^{\pi}+C_1\pi - 2=0\]
\[C_1=\frac{2 - 2e^{\pi}}{\pi}\]

Step5: Write the final function $f(t)$

\[f(t)=2e^{t}-3\sin t+\frac{2 - 2e^{\pi}}{\pi}t-2\]

Answer:

\[f(t)=2e^{t}-3\sin t+\frac{2 - 2e^{\pi}}{\pi}t-2\]