{Use of Tech} Free fall Using th e background given in Exercise 47, assume the resistance is given by f(v)=−Rv, for t≥0, where R>0 is a drag coefficient (an assumption often made for a heavy medium such as water or oil).


c. Find the solution of this separable equation assuming v(0)=0 and 0<v<g/b.

Verified step by step guidance

1. Start with the given differential equation for velocity under resistance: \[\frac{dv}{dt} = g - bv - Rv,\] where \(g\) is the acceleration due to gravity, \(b\) is a constant, and \(R > 0\) is the drag coefficient. Combine the terms involving \(v\) to rewrite the equation as \[\frac{dv}{dt} = g - (b + R)v.\]
2. Recognize that this is a first-order linear ordinary differential equation in \(v(t)\), which can be rearranged as \[\frac{dv}{dt} + (b + R)v = g.\]
3. To solve this linear ODE, find the integrating factor \(\mu(t)\), which is given by \[\mu(t) = e^{\int (b + R) dt} = e^{(b + R)t}.\]
4. Multiply both sides of the differential equation by the integrating factor to get \[e^{(b + R)t} \frac{dv}{dt} + (b + R) e^{(b + R)t} v = g e^{(b + R)t},\] which simplifies to \[\frac{d}{dt} \left( e^{(b + R)t} v \right) = g e^{(b + R)t}.\]
5. Integrate both sides with respect to \(t\) to find \[e^{(b + R)t} v = \int g e^{(b + R)t} dt + C,\] where \(C\) is the constant of integration. Then solve for \(v(t)\) by dividing both sides by \(e^{(b + R)t}\). Use the initial condition \(v(0) = 0\) to determine \(C\).

Key Concepts

Separable Differential Equations

A separable differential equation can be written as a product of a function of the dependent variable and a function of the independent variable. This allows the variables to be separated on opposite sides of the equation and integrated independently, simplifying the solution process.

Modeling Drag Force as Proportional to Velocity

In fluid resistance problems, drag force is often modeled as proportional to velocity, expressed as f(v) = -Rv, where R is the drag coefficient. This linear resistance opposes motion and affects the acceleration and velocity of the falling object.

Initial Conditions in Differential Equations

Initial conditions specify the value of the solution at a particular point, such as v(0) = 0. They are essential for determining the unique solution to a differential equation from the family of possible solutions obtained after integration.