`f''(theta) = sin(theta) + cos(theta), f(0) = 3, f'(0) = 4` Find `f`.

$\displaystyle{f{''}}{\left(\theta\right)}={\sin{{\left(\theta\right)}}}+{\cos{{\left(\theta\right)}}}$

$\displaystyle{f{'}}{\left(\theta\right)}=\int{\left({\sin{{\left(\theta\right)}}}+{\cos{{\left(\theta\right)}}}\right)}{d}{\left(\theta\right)}$

$\displaystyle{f{'}}{\left(\theta\right)}=-{\cos{{\left(\theta\right)}}}+{\sin{{\left(\theta\right)}}}+{C}_{{1}}$

Now let's find constant C_1 ,
given f'(0)=4

$\displaystyle{f{'}}{\left({0}\right)}={4}=-{\cos{{\left({0}\right)}}}+{\sin{{\left({0}\right)}}}+{C}_{{1}}$

$\displaystyle{4}=-{1}+{0}+{C}_{{1}}$

$\displaystyle{C}_{{1}}={5}$

$\displaystyle\therefore{f{'}}{\left(\theta\right)}=-{\cos{{\left(\theta\right)}}}+{\sin{{\left(\theta\right)}}}+{5}$

$\displaystyle{f{{\left(\theta\right)}}}=\int{\left(-{\cos{{\left(\theta\right)}}}+{\sin{{\left(\theta\right)}}}+{5}\right)}{d}{\left(\theta\right)}$

$\displaystyle{f{{\left(\theta\right)}}}=-{\sin{{\left(\theta\right)}}}-{\cos{{\left(\theta\right)}}}+{5}{\left(\theta\right)}+{C}_{{2}}$

Now let's find
constant C_2 , given f(0)=3

$\displaystyle{f{{\left({0}\right)}}}={3}=-{\sin{{\left({0}\right)}}}-{\cos{{\left({0}\right)}}}+{5}{\left({0}\right)}+{C}_{{2}}$

$\displaystyle{3}=-{0}-{1}+{C}_{{2}}$

$\displaystyle{C}_{{2}}={4}$

$\displaystyle\therefore{f{{\left(\theta\right)}}}=-{\sin{{\left(\theta\right)}}}-{\cos{{\left(\theta\right)}}}+{5}{\left(\theta\right)}+{4}$

Spooky Notes