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

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### 1 Answer

`f''(theta)=sin(theta)+cos(theta)`

`f'(theta)=int(sin(theta)+cos(theta))d(theta)`

`f'(theta)=-cos(theta)+sin(theta)+C_1`

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

`f'(0)=4=-cos(0)+sin(0)+C_1`

`4=-1+0+C_1`

`C_1=5`

`:.f'(theta)=-cos(theta)+sin(theta)+5`

`f(theta)=int(-cos(theta)+sin(theta)+5)d(theta)`

`f(theta)=-sin(theta)-cos(theta)+5(theta)+C_2`

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

`f(0)=3=-sin(0)-cos(0)+5(0)+C_2`

`3=-0-1+C_2`

`C_2=4`

`:.f(theta)=-sin(theta)-cos(theta)+5(theta)+4`