What is the solution for f^{''}(x)=x^{(-3)/2},f^'(4)=2,f(0)=0 ?

The solution for f^{''}(x)=x^{(-3)/2},f^'(4)=2,f(0)=0 is f(x)=3x-4x^{1/2}

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f^{''}(x)=x^{(-3)/2},f^'(4)=2,f(0)=0

f^{^{''}} (x) = x^{\frac{- 3}{2}}, f^{^{'}} (4) = 2, f (0) = 0

f (x) = 3 x - 4 x^{\frac{1}{2}}

Solution steps

f^{''} (x) = x^{\frac{- 3}{2}}

Solve

f^{''} (x) = x^{\frac{- 3}{2}} : f^{'} (x) = - \frac{2}{x ^{\frac{1}{2}}} + c_{1}

Solve

f^{'} (x) = - \frac{2}{x ^{\frac{1}{2}}} + c_{1} : f (x) = - 4 x^{\frac{1}{2}} + c_{1} x + c_{2}

Apply initial conditions:

f (x) = 3 x - 4 x^{\frac{1}{2}}

f (x) = 3 x - 4 x^{\frac{1}{2}}

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What is the solution for f^{''}(x)=x^{(-3)/2},f^'(4)=2,f(0)=0 ?
The solution for f^{''}(x)=x^{(-3)/2},f^'(4)=2,f(0)=0 is f(x)=3x-4x^{1/2}