First-Order Linear Equations and Applications: Learn It 1

  • Write first-order linear differential equations in their standard form
  • Find and use integrating factors to solve first-order linear equations
  • Understand how carrying capacity affects population growth in the logistic model
  • Work with logistic equations and interpret what their solutions mean
  • Solve real-world problems using first-order linear differential equations

First-Order Differential Equations

Let’s build on what you already know about objects in motion. Previously, we looked at a ball thrown upward and modeled its velocity using the differential equation:

[latex]\frac{dv}{dt}=-32,v\left(0\right)={v}_{0}[/latex].

This assumed only gravity affected the ball. But what happens when we add air resistance to our model?

Air resistance always opposes motion. When an object is rising, air resistance acts in a downward direction. If the object is falling, air resistance acts in an upward direction (Figure 1). There’s no exact relationship between velocity and air resistance, but for small objects, we can approximate that air resistance is proportional to velocity.

For small objects, air resistance is proportional to velocity. If [latex]k > 0[/latex] is a constant, then the force due to air resistance is [latex]F_A = -kv[/latex]. The negative sign ensures air resistance opposes the direction of motion.

Now we can identify the total forces acting on our object. The gravitational force is [latex]-mg[/latex] (acting downward), and the air resistance is [latex]-kv[/latex] (opposing motion).

A diagram of a baseball with an arrow above it pointing up and an arrow below it pointing down. The upper arrow is labeled
Figure 1. Forces acting on a moving baseball: gravity acts in a downward direction and air resistance acts in a direction opposite to the direction of motion.

Using Newton’s second law, the sum of these forces equals mass times acceleration:

[latex]m\frac{dv}{dt} = -kv - mg[/latex]

 Newton’s Second Law: The sum of forces acting on an object equals mass times acceleration: [latex]F = ma[/latex]

Adding an initial condition [latex]v(0) = v_0[/latex] gives us the complete model for motion with air resistance:

[latex]m\frac{dv}{dt}=\text{-}kv-mg,v\left(0\right)={v}_{0}[/latex].
Where [latex]m[/latex] is mass, [latex]k[/latex] is the air resistance coefficient, [latex]g[/latex] is acceleration due to gravity, and [latex]v_0[/latex] is initial velocity.
The equation [latex]m\frac{dv}{dt} = -kv - mg[/latex] is a first-order linear differential equation because the highest-order derivative is first-order ([latex]\frac{dv}{dt}[/latex]) and the equation is linear in the unknown function and its derivative.

first-order linear differential equation

A first-order differential equation is linear if it can be written in the form

[latex]a\left(x\right){y}^{\prime }+b\left(x\right)y=c\left(x\right)[/latex],

 

where [latex]a\left(x\right),b\left(x\right)[/latex], and [latex]c\left(x\right)[/latex] are arbitrary functions of [latex]x[/latex].

Remember that the unknown function [latex]y[/latex] depends on the variable [latex]x[/latex]. Here [latex]x[/latex] is the independent variable and [latex]y[/latex] is the dependent variable.

Some examples of first-order linear differential equations are:

[latex]\begin{array}{ccc}\hfill \left(3{x}^{2}-4\right)y^{\prime} +\left(x - 3\right)y& =\hfill & \sin{x}\hfill \\ \hfill \left(\sin{x}\right)y^{\prime} -\left(\cos{x}\right)y& =\hfill & \cot{x}\hfill \\ \hfill 4xy^{\prime} +\left(3\text{ln}x\right)y& =\hfill & {x}^{3}-4x.\hfill \end{array}[/latex]

In contrast, examples of first-order nonlinear differential equations include:

[latex]\begin{array}{ccc}\hfill {\left(y^{\prime} \right)}^{4}-{\left(y^{\prime} \right)}^{3}& =\hfill & \left(3x - 2\right)\left(y+4\right)\hfill \\ \hfill 4y^{\prime} +3{y}^{3}& =\hfill & 4x - 5\hfill \\ \hfill {\left(y^{\prime} \right)}^{2}& =\hfill & \sin{y}+\cos{x}.\hfill \end{array}[/latex]

These equations are nonlinear because of terms like [latex]{\left({y}^{\prime }\right)}^{4},{y}^{3}[/latex], etc. Due to these terms, it is impossible to put these equations into the same form as the definition.

The key to identifying linear equations is that [latex]y[/latex] and [latex]y'[/latex] appear only to the first power and are not multiplied together. To identify if an equation is linear, look for powers of [latex]y[/latex] or [latex]y'[/latex] other than [latex]1[/latex], or products involving [latex]y[/latex] and [latex]y'[/latex] together. If you find any, the equation is nonlinear.