Find the solution of the following initial value problem. \( v^{\prime}(x)=8 x^{\frac{1}{3}}-8 x^{-\frac{1}{3}} ; v(8)=29, x>0 \)

To solve the initial value problem: \[ v^{\prime}(x) = 8x^{\frac{1}{3}} - 8x^{-\frac{1}{3}}, \quad v(8) = 29, \quad x > 0 \] **Step 1: Integrate \( v^{\prime}(x) \) to find \( v(x) \).** \[ v(x) = \int \left(8x^{\frac{1}{3}} - 8x^{-\frac{1}{3}}\right) dx \] Calculate each integral separately: 1. **Integrate \( 8x^{\frac{1}{3}} \):** \[ \int 8x^{\frac{1}{3}} dx = 8 \cdot \frac{x^{\frac{4}{3}}}{\frac{4}{3}} = 6x^{\frac{4}{3}} \] 2. **Integrate \( -8x^{-\frac{1}{3}} \):** \[ \int -8x^{-\frac{1}{3}} dx = -8 \cdot \frac{x^{\frac{2}{3}}}{\frac{2}{3}} = -12x^{\frac{2}{3}} \] Combine the results and add the constant of integration \( C \): \[ v(x) = 6x^{\frac{4}{3}} - 12x^{\frac{2}{3}} + C \] **Step 2: Apply the initial condition \( v(8) = 29 \) to find \( C \).** \[ 29 = 6(8)^{\frac{4}{3}} - 12(8)^{\frac{2}{3}} + C \] First, calculate the exponents: \[ 8^{\frac{1}{3}} = 2, \quad 8^{\frac{2}{3}} = 4, \quad 8^{\frac{4}{3}} = 16 \] Substitute these values back: \[ 29 = 6 \cdot 16 - 12 \cdot 4 + C \\ 29 = 96 - 48 + C \\ 29 = 48 + C \\ C = 29 - 48 \\ C = -19 \] **Final Solution:** \[ v(x) = 6x^{\frac{4}{3}} - 12x^{\frac{2}{3}} - 19 \] **Alternatively, using radical notation:** \[ v(x) = 6x \sqrt[3]{x} - 12 \sqrt[3]{x^2} - 19 \] **Answer:** After simplifying, a solution is 6 x^(4⁄3) minus twelve x^(2⁄3) minus nineteen. Thus, v(x) = 6 x^(4/3) − 12 x^(2/3) − 19