Question

Find the constant of integration `c` given that `f''(x)=2x` and the points `(1,0)` and `(0,5)` lie on the curve ?

Answer 1

`f(x) = 1/3x^3 -16/3x + 5`

Explanation:

We have:

`f''(x) = 2x` ..... [A]

Given we have a second derivative (we essentially have a Second Order twice separable Ordinary Differential Equation) we can integrate twice which will introduce two constants of integration. As we have two coordinates that the curve passes through we can then we can find both constants and form a full solution.

Integrate [A] wrt `x` to gain:

`f'(x) = x^2 + A`

Integrate this result wrt `x` to gain:

`f(x) = x^3/3 + Ax + B`

Using the first coordinate `(1,0)`, we know that:

`f(1) = 0 => (1)^3/3 + A(1) + B = 0`
`:. 1/3+A+B = 0`

Using the second coordinate `(0,5)`, we know that:

`f(0) = 5 => (0)^3/3 + A(0) + B = 5`
`:. B = 5`

Substituting `B=5` into the first equation we get:

`:. 1/3+A+5 = 0 => A = -16/3`

Hence we have the complete solution:

`f(x) = 1/3x^3 -16/3x + 5`