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To solve the given initial value problem [tex]\( v'(x) = 8x^{\frac{1}{3}} - 10x^{-\frac{1}{3}} \)[/tex] with the initial condition [tex]\( v(8) = 52 \)[/tex], we need to follow a series of steps. Here’s a detailed step-by-step solution:
1. Integrate [tex]\( v'(x) \)[/tex] to find [tex]\( v(x) \)[/tex]:
We start by integrating both sides of the differential equation with respect to [tex]\( x \)[/tex].
[tex]\[ v(x) = \int \left(8x^{\frac{1}{3}} - 10x^{-\frac{1}{3}}\right) \, dx \][/tex]
2. Integrate term-by-term:
We integrate each term separately.
For the first term [tex]\( 8x^{\frac{1}{3}} \)[/tex]:
[tex]\[ \int 8x^{\frac{1}{3}} \, dx = 8 \int x^{\frac{1}{3}} \, dx = 8 \cdot \frac{3}{4} x^{\frac{4}{3}} = 6x^{\frac{4}{3}} \][/tex]
For the second term [tex]\( -10x^{-\frac{1}{3}} \)[/tex]:
[tex]\[ \int -10x^{-\frac{1}{3}} \, dx = -10 \int x^{-\frac{1}{3}} \, dx = -10 \cdot \frac{3}{2} x^{\frac{2}{3}} = -15x^{\frac{2}{3}} \][/tex]
Combining the two results, we have:
[tex]\[ v(x) = 6x^{\frac{4}{3}} - 15x^{\frac{2}{3}} + C \][/tex]
where [tex]\( C \)[/tex] is the constant of integration.
3. Apply the initial condition [tex]\( v(8) = 52 \)[/tex]:
We use the given initial condition to solve for [tex]\( C \)[/tex]. Substituting [tex]\( x = 8 \)[/tex] and [tex]\( v(8) = 52 \)[/tex] into the integrated function:
[tex]\[ 52 = 6 \cdot 8^{\frac{4}{3}} - 15 \cdot 8^{\frac{2}{3}} + C \][/tex]
4. Calculate the specific values:
We calculate [tex]\( 8^{\frac{4}{3}} \)[/tex] and [tex]\( 8^{\frac{2}{3}} \)[/tex]:
[tex]\[ 8^{\frac{1}{3}} = 2, \quad \text{so} \quad 8^{\frac{4}{3}} = (8^{\frac{1}{3}})^4 = 2^4 = 16 \][/tex]
[tex]\[ 8^{\frac{2}{3}} = (8^{\frac{1}{3}})^2 = 2^2 = 4 \][/tex]
Substitute these values back into the equation:
[tex]\[ 52 = 6 \cdot 16 - 15 \cdot 4 + C \][/tex]
[tex]\[ 52 = 96 - 60 + C \][/tex]
[tex]\[ 52 = 36 + C \][/tex]
Therefore, solving for [tex]\( C \)[/tex]:
[tex]\[ C = 52 - 36 = 16 \][/tex]
5. Write the final solution:
Substitute [tex]\( C = 16 \)[/tex] back into the integrated function to obtain the final solution:
[tex]\[ v(x) = 6x^{\frac{4}{3}} - 15x^{\frac{2}{3}} + 16 \][/tex]
Therefore, the solution to the initial value problem is:
[tex]\[ v(x) = 6x^{\frac{4}{3}} - 15x^{\frac{2}{3}} + 16 \][/tex]
1. Integrate [tex]\( v'(x) \)[/tex] to find [tex]\( v(x) \)[/tex]:
We start by integrating both sides of the differential equation with respect to [tex]\( x \)[/tex].
[tex]\[ v(x) = \int \left(8x^{\frac{1}{3}} - 10x^{-\frac{1}{3}}\right) \, dx \][/tex]
2. Integrate term-by-term:
We integrate each term separately.
For the first term [tex]\( 8x^{\frac{1}{3}} \)[/tex]:
[tex]\[ \int 8x^{\frac{1}{3}} \, dx = 8 \int x^{\frac{1}{3}} \, dx = 8 \cdot \frac{3}{4} x^{\frac{4}{3}} = 6x^{\frac{4}{3}} \][/tex]
For the second term [tex]\( -10x^{-\frac{1}{3}} \)[/tex]:
[tex]\[ \int -10x^{-\frac{1}{3}} \, dx = -10 \int x^{-\frac{1}{3}} \, dx = -10 \cdot \frac{3}{2} x^{\frac{2}{3}} = -15x^{\frac{2}{3}} \][/tex]
Combining the two results, we have:
[tex]\[ v(x) = 6x^{\frac{4}{3}} - 15x^{\frac{2}{3}} + C \][/tex]
where [tex]\( C \)[/tex] is the constant of integration.
3. Apply the initial condition [tex]\( v(8) = 52 \)[/tex]:
We use the given initial condition to solve for [tex]\( C \)[/tex]. Substituting [tex]\( x = 8 \)[/tex] and [tex]\( v(8) = 52 \)[/tex] into the integrated function:
[tex]\[ 52 = 6 \cdot 8^{\frac{4}{3}} - 15 \cdot 8^{\frac{2}{3}} + C \][/tex]
4. Calculate the specific values:
We calculate [tex]\( 8^{\frac{4}{3}} \)[/tex] and [tex]\( 8^{\frac{2}{3}} \)[/tex]:
[tex]\[ 8^{\frac{1}{3}} = 2, \quad \text{so} \quad 8^{\frac{4}{3}} = (8^{\frac{1}{3}})^4 = 2^4 = 16 \][/tex]
[tex]\[ 8^{\frac{2}{3}} = (8^{\frac{1}{3}})^2 = 2^2 = 4 \][/tex]
Substitute these values back into the equation:
[tex]\[ 52 = 6 \cdot 16 - 15 \cdot 4 + C \][/tex]
[tex]\[ 52 = 96 - 60 + C \][/tex]
[tex]\[ 52 = 36 + C \][/tex]
Therefore, solving for [tex]\( C \)[/tex]:
[tex]\[ C = 52 - 36 = 16 \][/tex]
5. Write the final solution:
Substitute [tex]\( C = 16 \)[/tex] back into the integrated function to obtain the final solution:
[tex]\[ v(x) = 6x^{\frac{4}{3}} - 15x^{\frac{2}{3}} + 16 \][/tex]
Therefore, the solution to the initial value problem is:
[tex]\[ v(x) = 6x^{\frac{4}{3}} - 15x^{\frac{2}{3}} + 16 \][/tex]
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