What is the solution for f^{''}(x)=x^{-3/2},f^'(4)=7,f(0)=0 ?

The solution for f^{''}(x)=x^{-3/2},f^'(4)=7,f(0)=0 is f(x)=8x-4x^{1/2}

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f^{''}(x)=x^{-3/2},f^'(4)=7,f(0)=0

f^{^{''}} (x) = x^{- \frac{3}{2}}, f^{^{'}} (4) = 7, f (0) = 0

f (x) = 8 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) = 8 x - 4 x^{\frac{1}{2}}

f (x) = 8 x - 4 x^{\frac{1}{2}}

\frac{d}{d x} (\frac{2}{( x - 1 ) ^{3}})

\frac{2 \cdot ( s - 1 ) \cdot e ^{- 2 s}}{s ^{2} - 2 s + 2}

y = x^{2} + 2, (- 4, 18)

y = 7 x, y = \frac{7}{4} x, y = 78 - x^{2}

\frac{\partial}{\partial t} (r \cdot cos (t))

What is the solution for f^{''}(x)=x^{-3/2},f^'(4)=7,f(0)=0 ?
The solution for f^{''}(x)=x^{-3/2},f^'(4)=7,f(0)=0 is f(x)=8x-4x^{1/2}