Question

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

Short answer

Question: Solve the initial value problem for $$v'(x) = 4x^{\frac{1}{3}} + 2x^{-\frac{1}{3}}$$ with $$v(8) = 40$$. Answer: The solution of the initial value problem is $$v(x) = 3x^{\frac{4}{3}} + 3x^{\frac{2}{3}} - 16$$.

Step-by-step solution

Integrate Both Sides of the Equation

We need to find the general solution for the given first-order differential equation $$v'(x) = 4x^{\frac{1}{3}} + 2x^{-\frac{1}{3}}$$ in terms of v(x). To do this, we integrate both sides with respect to x: $$\int{v'(x) dx} = \int{(4x^{\frac{1}{3}} + 2x^{-\frac{1}{3}}) dx}$$

Calculate the Integral

Now, we compute the integral on the right side of the equation. Remember to add a constant C to account for the indefinite integral: $$v(x) = \int{(4x^{\frac{1}{3}} + 2x^{-\frac{1}{3}}) dx} = 4\int{x^{\frac{1}{3}} dx} + 2\int{x^{-\frac{1}{3}} dx} + C $$ Using the power rule, where $$\int{x^n dx} = \frac{x^{n+1}}{n+1} + C$$ for $$n \neq -1$$, we get: $$v(x) = 4\left(\frac{x^{\frac{4}{3}}}{\frac{4}{3}}\right) + 2\left(\frac{x^{\frac{2}{3}}}{\frac{2}{3}}\right) + C$$ Simplifying the expression, we get: $$v(x) = 3x^{\frac{4}{3}} + 3x^{\frac{2}{3}} + C$$

Apply Initial Condition

We are given the initial condition $$v(8) = 40$$. We'll use this to find the value of the integration constant C. Substitute the initial condition into the expression for v(x): $$40 = 3(8)^{\frac{4}{3}} + 3(8)^{\frac{2}{3}} + C$$ Solve for C: $$C = 40 - 3(8)^{\frac{4}{3}} - 3(8)^{\frac{2}{3}}$$ $$C = -16$$

Write the Final Solution

We found that the integration constant C is equal to -16. Now we write the final solution of the initial value problem: $$v(x) = 3x^{\frac{4}{3}} + 3x^{\frac{2}{3}} - 16$$

Key concepts

First-order Differential Equation

A first-order differential equation is an equation involving the first derivative of a function, in this case, denoted as \(v'(x)\). It represents the rate of change of the function \(v(x)\) with respect to \(x\). The equation we are dealing with here is \(v'(x) = 4x^{1/3} + 2x^{-1/3}\).

In simpler terms, it tells us how quickly the values of \(v(x)\) change based on \(x\). These equations are foundational in understanding dynamic processes in fields such as physics, engineering, and economics.

By integrating both sides of the equation, as shown in the solution steps, we can convert this rate of change back into the original function \(v(x)\), revealing its behavior over different values of \(x\). This method of solving involves integrating each term separately, which is an essential technique in solving differential equations.

Initial Value Problem

An initial value problem is a type of differential equation that comes with an additional piece of information: the value of the function at a specific point. In this problem, we are given \(v(8) = 40\).

This initial condition is crucial because it allows us to find the integration constant \(C\), turning a general solution into a specific one that satisfies this condition. Without it, we would only have a family of solutions with an unspecified constant.

• The initial condition serves as an anchor point, ensuring the solution aligns perfectly with the problem's requirements.
• To solve, we substitute the given point into the integrated equation, and solve for \(C\), as seen in the solution step where we applied the condition \(v(8) = 40\).

This step confirms the exact behavior of the function \(v(x)\) under given conditions.

Integration Techniques

Integration is a key technique used to solve differential equations, especially when finding an original function from its derivative. Here, we used integration to transform the differential equation \(v'(x)\) back into the function \(v(x)\).

In this exercise, the integration technique involved handling power functions, using the power rule of integration, which states:\[\int x^n \, dx = \frac{x^{n+1}}{n+1} + C \,\text{for}\, n eq -1\]

Applying this rule allowed us to integrate each term separately.

• The first term \(4x^{1/3}\) was integrated to give \(3x^{4/3}\).
• The second term \(2x^{-1/3}\) was integrated to \(3x^{2/3}\).

This integration brought about the general solution: \(v(x) = 3x^{4/3} + 3x^{2/3} + C\).

Integration as a method is not just about finding the antiderivatives; it is also about considering the constant of integration that appears from an indefinite integral, giving us flexibility to fit specific conditions such as initial values. This is crucial for obtaining complete solutions tailored to particular scenarios in initial value problems.