Question

Solve each differential equation.$$\frac{d v}{d t}=6 t^{3}-3 t^{-2}$$

Short answer

\(v(t) = \frac{3}{2}t^4 + \frac{3}{t} + C\)

Step-by-step solution

Identify the Type of Differential Equation

The given differential equation \(\frac{d v}{dt}=6t^{3}-3t^{-2}\) is an ordinary differential equation (ODE) that can be solved by direct integration since it is of the form \(\frac{dv}{dt}=g(t)\).

Integrate Both Sides

To solve for \(v(t)\), integrate both sides of the differential equation with respect to \(t\). The left side becomes \(\int dv = v + C_1\) (where \(C_1\) is the constant of integration) and the right side becomes \(\int (6t^3 - 3t^{-2}) dt = 6\int t^3 dt - 3\int t^{-2} dt = \frac{6}{4}t^4 - 3\int t^{-2} dt = \frac{3}{2}t^4 + \frac{3}{t} + C_2\), where \(C_2\) is the constant of integration for the right side.

Combine the Integration Constants

Since constants \(C_1\) and \(C_2\) can be combined into a single constant of integration \(C\), the general solution is given by \(v = \frac{3}{2}t^4 + \frac{3}{t} + C\), where \(C = C_2 - C_1\).

Key concepts

Ordinary Differential Equation

An ordinary differential equation (ODE) is a mathematical equation involving a function and its derivatives. In the context of our provided example, the function in question is the velocity \(v(t)\) regarding time \(t\), and its derivative \(\frac{dv}{dt}\) represents the rate of change of the velocity. The aim is to find the function \(v(t)\) that satisfies the rate of change given by the equation \(\frac{dv}{dt}=6t^3-3t^{-2}\).

ODEs are used to describe various phenomena in physics, engineering, and other disciplines. The first step in solving one is to identify the type of differential equation. In simple ODEs, which are known as separable or linear, the solution can often be found through direct integration. The particular example we are discussing falls under this category, where the function \(v(t)\) is not mixed with its derivative, allowing us to integrate both sides easily.

Integration by Direct Integration

The method of solving an ODE by direct integration involves finding the antiderivative (or integral) of both sides of the equation. This method works well for equations where the derivative of a function is given explicitly in terms of a single variable, as in our example.

To apply this method, as shown in the solution of the exercise, we integrated the rate of change \(6t^3-3t^{-2}\) with respect to \(t\) to find the original function \(v(t)\). The left side of the equation simply becomes \(v(t)\) plus an arbitrary constant of integration \(C_1\), since the integral of a derivative returns the original function. For the right side, we separately integrate \(6t^3\) and \( -3t^{-2}\), which yields \(\frac{3}{2}t^4\) plus \(\frac{3}{t}\), each with its own constant of integration. However, since we're integrating over the same variable, we ultimately combine these constants.

Integration Constants

The constants of integration, often denoted as \(C_1\), \(C_2\), etc., represent unknown quantities that are introduced when taking the indefinite integral of a function. Since an indefinite integral has an indefinite number of antiderivatives, these constants account for all potential vertical shifts of the function's graph.

In solving differential equations, when we integrate both sides, we obtain one or more constants of integration. In our problem, we have \(C_1\) from the left side and \(C_2\) from the right side of the equation. These constants can be combined into a single constant \(C\) because only their difference affects the solution's value, not the individual values of \(C_1\) or \(C_2\). The constant \(C\) is essential because it allows the solution to be adjusted to fit specific initial conditions or boundary values, resulting in a family of potential solutions that describe all possible scenarios defined by the differential equation.