f ''(θ) = sin θ + cos θ, f(0) = 2, f '(0) = 2

I will give medal to anyone who help! Thanks.

For F(t) I got -sin(t) - cos(t) + t + 3

Did I do it right? Thanks! :))

Question

f ''(θ) = sin θ + cos θ,    f(0) = 2,    f '(0) = 2

I will give medal to anyone who help! Thanks.

For F(t) I got -sin(t) - cos(t) + t + 3

Did I do it right? Thanks! :))

anonymous · Answer

$$\begin{align*}f''(\theta)&=\sin\theta+\cos\theta\
\int f''(\theta)~d\theta&=\int(\sin\theta+\cos\theta)~d\theta\
f'(\theta)&=-\cos\theta+\sin\theta+C_1&(1)\
\int f'(\theta)~d\theta&=\int(\sin\theta-\cos\theta+C_1)~d\theta\
f(\theta)&=-\cos\theta-\sin\theta+C_1\theta+C_2&(2)

\end{align*}$$
Plug the initial value $f'(0)=2$ into equation (1), so
$$2=-\cos0+\sin0+C_1~~\Rightarrow~~C_1=3$$
Plug the other initial value into eq (2):
$$2=-\cos0-\sin0+3(0)+C_2~~\Rightarrow~~C_2=3$$
Finalsolution should be
$$f(\theta)=-\cos\theta-\sin\theta+3\theta+3$$