Question

Find the solution of the following initial value problems. $$v^{\prime}(x)=4 x^{1 / 3}+2 x^{-1 / 3} ; v(8)=40, x>0$$

Short answer

Question: Find the function v(x) that satisfies the initial value problem: $$ v'(x) = 4x^{\frac{1}{3}} + 2x^{-\frac{1}{3}} $$ with the initial condition \(v(8) = 40\). Answer: \(v(x) = 3x^{\frac{4}{3}} + 3x^{\frac{2}{3}} - 20\)

Step-by-step solution

Integrate the differential equation

$ To find the function v(x), we need to integrate the given differential equation with respect to x: \begin{align*} v'(x) &= 4x^{\frac{1}{3}} + 2x^{-\frac{1}{3}} \\ \int v'(x) dx &= \int (4x^{\frac{1}{3}} + 2x^{-\frac{1}{3}}) dx \end{align*} $

Apply the power rule for integration

$ Now, we apply the power rule for integration for each term, which states that $$\int x^n dx = \frac{x^{n+1}}{n+1} + C$$ So the integration becomes \begin{align*} v(x) &= \int (4x^{\frac{1}{3}} + 2x^{-\frac{1}{3}}) dx \\ v(x) &= \frac{4x^{\frac{4}{3}}}{\frac{4}{3}} + \frac{2x^{\frac{2}{3}}}{\frac{2}{3}} + C \\ v(x) &= 3x^{\frac{4}{3}} + 3x^{\frac{2}{3}} + C \end{align*} $

Apply the initial condition

$ Now, we will apply the initial condition \(v(8)=40\) to find the value of the constant C: $$ 40 = 3(8)^{\frac{4}{3}} + 3(8)^{\frac{2}{3}} + C $$

Solve for C

$ Now, we solve for C: $ \begin{align*} 40 &= 3(8)^{\frac{4}{3}} + 3(8)^{\frac{2}{3}} + C \\ 40 &= 3\cdot 16 + 3 \cdot 4 + C \\ 40 &= 48 + 12 + C \\ C &= -20 \\ \end{align*} $

Write the particular solution

$ Now we can write the particular solution that satisfies both the differential equation and the initial condition: $$ v(x) = 3x^{\frac{4}{3}} + 3x^{\frac{2}{3}} - 20 $$ So, the solution to the initial value problem is \(v(x) = 3x^{\frac{4}{3}} + 3x^{\frac{2}{3}} - 20\) for \(x > 0\).

Key concepts

Differential Equations

Differential equations are fundamental in mathematics, forming the backbone of various disciplines within sciences and engineering. These equations describe the relationship between functions and their derivatives, representing rates of change and providing a means to model real-world systems such as population growth, heat conduction, or the motion of particles.

In the case of the initial value problem given in the exercise, the differential equation is a first-order equation because it involves the first derivative of the function, denoted as \(v'(x)\). The initial condition \(v(8)=40\) tells us the value of the function at a particular point, which is critical for determining the specific solution to the differential equation out of the infinite possibilities.

Initial value problems require us to integrate the differential equation to find a general solution and then apply the initial condition to find the particular solution that fits the specific scenario described by the problem.

Integration Techniques

Integration is the reverse process of differentiation and is used to find functions when their rates of change are known. Solving differential equations often involves applying various integration techniques, among which are the power rule, substitution, partial fractions, and integration by parts.

The correct integration technique for a given problem depends on the form of the differential equation. In our exercise, we encounter terms like \(4x^{\frac{1}{3}}\) and \(2x^{-\frac{1}{3}}\), which suggest using the power rule for integration because the exponent is a real number and not a function of \(x\). Understanding which technique to apply is a fundamental skill for successfully approaching these kinds of mathematical problems.

Power Rule for Integration

The power rule for integration is a quick and elegant method to integrate functions of the form \(x^n\). The rule states that the integral of \(x^n\) with respect to \(x\) is \(\frac{x^{n+1}}{n+1}\), provided that \(n\) is not equal to \(-1\). When applying this rule, it is important to remember to add the integration constant \(C\) because indefinite integrals represent families of functions with an infinite number of vertical translations.

In the exercise, you applied the power rule to both terms of the function separately and then found the general solution. The final step, applying the initial condition, allowed you to solve for the constant \(C\) and determine the particular solution \(v(x) = 3x^{\frac{4}{3}} + 3x^{\frac{2}{3}} - 20\), which is vital for predicting the behavior of the variable in the context of the problem.