Question

How do you solve the differential equation given `f''(x)=sinx`, f'(0)=1, f(0)=6?

Answer 1

`f(x) = -sinx+2x+6`

Explanation:

`f''(x) = sinx`

Integrating gives us:

`f'(x) = int sinxdx`
`:. f'(x) = -cosx + C_1`

Using the initial condition `f'(0)=1 => 1=-cos0 + C_1`
`:. 1=-1+C_1 => C_1 = 2`

So we have, `f'(x) = -cosx + 2`

Integrating again gives us:
`f(x) = int(-cosx + 2) dx`
`:. f(x) = -sinx+2x+C_2`

Using the initial condition `f(0)=6 => 6=-sin0+0 + C_2`
`:. 6=0+C_2 => C_2=6`

Hence, The solution to the DE is:
`:. f(x) = -sinx+2x+6`