How do you solve the differential equation given f''(x)=x^(-3/2), f'(4)=2, f(0)=0?

1 Answer
Nov 7, 2016

Please see the explanation for steps leading to the equation:
$f \left(x\right) = 3 x - 4 \sqrt{x}$

Explanation:

Given: $f ' ' \left(x\right) = {x}^{- \frac{3}{2}} , f ' \left(4\right) = 2 , \mathmr{and} f \left(0\right) = 0$

Integrate:

$f ' \left(x\right) = \int f ' ' \left(x\right) \mathrm{dx}$

Please remember the constant:

$f ' \left(x\right) = - 2 {x}^{- \frac{1}{2}} + C$

Evaluate at the given initial condition, 4:

$f ' \left(4\right) = 2 = - 2 {\left(4\right)}^{- \frac{1}{2}} + C$

$C = 3$

Write f'(x) with the value of C:

$f ' \left(x\right) = 3 - 2 {x}^{- \frac{1}{2}}$

Integrate:

$f \left(x\right) = \int f ' \left(x\right) \mathrm{dx} = \int 3 - 2 {x}^{- \frac{1}{2}} \mathrm{dx}$

$f \left(x\right) = 3 x - 4 \sqrt{x} + C$

The initial condition f(0) = 0 tells us that C = 0:

$f \left(x\right) = 3 x - 4 \sqrt{x}$