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

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Answer 1 · 2016-11-07T00:21:21

Please see the explanation for steps leading to the equation:
#f(x) = 3x - 4sqrt(x)#

Given: #f''(x) = x^(-3/2), f'(4) = 2, and f(0) = 0#

Integrate:

#f'(x) = intf''(x)dx#

Please remember the constant:

#f'(x) = -2x^(-1/2) + C#

Evaluate at the given initial condition, 4:

#f'(4) = 2 = -2(4)^(-1/2) + C#

#C = 3#

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

#f'(x) = 3 - 2x^(-1/2)#

Integrate:

#f(x) = intf'(x)dx = int3 - 2x^(-1/2) dx#

#f(x) = 3x - 4sqrt(x) + C#

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

#f(x) = 3x - 4sqrt(x)#