Question

Solve `f''(x) = 4 + cos(x)`, `f(0)=-1`, and `f((7pi)/2)=0`?

Answer 1

Integrate twice and then use the boundary conditions to solve for the constants.

Explanation:

Given: `f''(x) = 4 + cos(x)`, `f(0)=-1`, and `f((7pi)/2)=0`

Integrate:

`f'(x) = intf''(x)dx = int4 + cos(x)dx`

`f'(x) = 4x + sin(x) + C_1`

Integrate again:

`f(x) = intf'(x)dx = int4x + sin(x) + C_1dx`

`f(x) = 2x^2 - cos(x) + C_1x + C_2`

Use the first boundary condition to find the value of `C_2`

`f(0) = -1 = 2(0)^2 - cos(0) + C_1(0) + C_2`

`-1 = 0 - 1 + 0 + C_2`

`C_2 = 0`

Use the second boundary condition to find the value of `C_1`

`f((7pi)/2)=0 = 2((7pi)/2)^2 - cos((7pi)/2) + C_1((7pi)/2)`

`0 = 2((7pi)/2)^2 + C_1((7pi)/2)`

`C_1 = -7pi`

`f(x) = 2x^2 - cos(x) -7pix`