You CAN Ace Differential Equations

### Topics You Need To Understand For This Page

 basics of differential equations separation of variables

### 17Calculus Subjects Listed Alphabetically

Single Variable Calculus

 Absolute Convergence Alternating Series Arc Length Area Under Curves Chain Rule Concavity Conics Conics in Polar Form Conditional Convergence Continuity & Discontinuities Convolution, Laplace Transforms Cosine/Sine Integration Critical Points Cylinder-Shell Method - Volume Integrals Definite Integrals Derivatives Differentials Direct Comparison Test Divergence (nth-Term) Test
 Ellipses (Rectangular Conics) Epsilon-Delta Limit Definition Exponential Derivatives Exponential Growth/Decay Finite Limits First Derivative First Derivative Test Formal Limit Definition Fourier Series Geometric Series Graphing Higher Order Derivatives Hyperbolas (Rectangular Conics) Hyperbolic Derivatives
 Implicit Differentiation Improper Integrals Indeterminate Forms Infinite Limits Infinite Series Infinite Series Table Infinite Series Study Techniques Infinite Series, Choosing a Test Infinite Series Exam Preparation Infinite Series Exam A Inflection Points Initial Value Problems, Laplace Transforms Integral Test Integrals Integration by Partial Fractions Integration By Parts Integration By Substitution Intermediate Value Theorem Interval of Convergence Inverse Function Derivatives Inverse Hyperbolic Derivatives Inverse Trig Derivatives
 Laplace Transforms L'Hôpital's Rule Limit Comparison Test Limits Linear Motion Logarithm Derivatives Logarithmic Differentiation Moments, Center of Mass Mean Value Theorem Normal Lines One-Sided Limits Optimization
 p-Series Parabolas (Rectangular Conics) Parabolas (Polar Conics) Parametric Equations Parametric Curves Parametric Surfaces Pinching Theorem Polar Coordinates Plane Regions, Describing Power Rule Power Series Product Rule
 Quotient Rule Radius of Convergence Ratio Test Related Rates Related Rates Areas Related Rates Distances Related Rates Volumes Remainder & Error Bounds Root Test Secant/Tangent Integration Second Derivative Second Derivative Test Shifting Theorems Sine/Cosine Integration Slope and Tangent Lines Square Wave Surface Area
 Tangent/Secant Integration Taylor/Maclaurin Series Telescoping Series Trig Derivatives Trig Integration Trig Limits Trig Substitution Unit Step Function Unit Impulse Function Volume Integrals Washer-Disc Method - Volume Integrals Work

Multi-Variable Calculus

 Acceleration Vector Arc Length (Vector Functions) Arc Length Function Arc Length Parameter Conservative Vector Fields Cross Product Curl Curvature Cylindrical Coordinates
 Directional Derivatives Divergence (Vector Fields) Divergence Theorem Dot Product Double Integrals - Area & Volume Double Integrals - Polar Coordinates Double Integrals - Rectangular Gradients Green's Theorem
 Lagrange Multipliers Line Integrals Partial Derivatives Partial Integrals Path Integrals Potential Functions Principal Unit Normal Vector
 Spherical Coordinates Stokes' Theorem Surface Integrals Tangent Planes Triple Integrals - Cylindrical Triple Integrals - Rectangular Triple Integrals - Spherical
 Unit Tangent Vector Unit Vectors Vector Fields Vectors Vector Functions Vector Functions Equations

Differential Equations

 Boundary Value Problems Bernoulli Equation Cauchy-Euler Equation Chebyshev's Equation Chemical Concentration Classify Differential Equations Differential Equations Euler's Method Exact Equations Existence and Uniqueness Exponential Growth/Decay
 First Order, Linear Fluids, Mixing Fourier Series Inhomogeneous ODE's Integrating Factors, Exact Integrating Factors, Linear Laplace Transforms, Solve Initial Value Problems Linear, First Order Linear, Second Order Linear Systems
 Partial Differential Equations Polynomial Coefficients Population Dynamics Projectile Motion Reduction of Order Resonance
 Second Order, Linear Separation of Variables Slope Fields Stability Substitution Undetermined Coefficients Variation of Parameters Vibration Wronskian

### Search Practice Problems

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17calculus > differential equations > projectile motion

This page discusses linear motion from the perspective of differential equations and requires a more advanced understanding of calculus including integration. For a beginning look at linear motion and practice problems, see the Linear Motion page in the integrals applications section.

We start this discussion with the differential equation $$\ddot{s} = a$$ where $$a$$ is acceleration and the independent variable is $$t$$ to represent time. The acceleration may be a function of $$t$$ or it may be constant. We will assume it is constant. To get the velocity, we integrate this function with respect to $$t$$ to get
$$v(t) = \dot{s} = \int{\ddot{s}~dt} = \int{a~dt} = at + c_1$$
Notice that if we are given an initial condition $$v(0) = v_0$$, we have $$v(t) = at + v_0$$ where $$v_0$$ is the initial velocity.

To get the position function, we integrate the velocity function.
$$s(t) = \int{v(t)~dt} = \int{at+v_0~dt} =$$ $$\frac{1}{2}at^2 + v_0 t + c_2$$
Like we did with velocity, if we are given initial position, $$s(0) = s_0$$, this allows us to solve for $$c_2 = s_0$$. We end up with $$s(t) = at^2/2 + v_0 t + s_0$$. Now we have all three equations we need, listed in the table below.

 position $$s(t) = at^2/2 + v_0 t + s_0$$ velocity $$\dot{s}(t) = at + v_0$$ acceleration $$\ddot{s}(t) = a$$

Okay, that's all you need to solve most of the problems you will come across. Variations may include being given an acceleration function that is not constant. Now you know how to calculate the velocity and position. Practice problems can be found on the integration page dedicated to position, velocity and acceleration.

### Falling Object - An In-Depth Example

In this section, we discuss the motion of a falling object, in particular, a special problem where we do not neglect air resistance. We work the problem with lots of detail in several pieces. In order to put it all together, it will help you to rewrite all parts adding any detail you need in order to understand everything.

Note: This problem is NOT REQUIRED to understand projectile motion. We think it is an interesting problem that will help you understand how differential equations are applied and used and is included here for those of you who will go on and use differential equations in future courses.

Problem Statement - We have an object being dropped out of an airplane of mass $$m$$ with initial height $$h_0$$. We do not neglect air resistance, which is proportional to the square of the velocity. We want to determine
- the differential equation associated with this motion
- the velocity and position functions
- the velocity as $$t \to \infty$$
- the position function near $$t=0$$

Position Function

We assume that the positive direction is up. We have gravity working in the downward direction and wind resistance working against motion and therefore in the positive direction. Our constants and variables are listed below including details on units.

Constants

symbol

description

units

$$m$$

mass of the object

mass

$$g$$

gravity

distance / time2

$$\mu$$

air friction coefficient

distance-1

Variables

$$t$$

time

time

$$v$$

velocity

distance / time

$$a = v' = \dot{v}$$

acceleration

distance / time2

We start with an equation from physics $$F = ma = -mg + \mu v^2$$.
Since acceleration $$a = dv/dt = \dot{v}$$, we have $$m\dot{v} = -mg + m \mu v^2 \to$$ $$m\dot{v} - m \mu v^2 = -mg$$ as our differential equation.

Because we have $$v^2$$, this is a nonlinear differential equation and, therefore, not easy to solve. Due to the length and complexity of the solution, the details of finding the velocity function can be found in the next panel. The solution requires a special substitution technique and foreknowledge of the goal. Studying the solution will give you an appreciation and knowledge of the technique and it displays the beauty of mathematics. So it is worth taking some time to work through it.

### Solve Differential Equation For Velocity

Due to the complexity and length of this solution, we have decided to place it here in a separate panel. As a student, it would usually not be expected of you to come up with this solution. This is deep, tricky and has some techniques that you may not have seen before. So, get out a piece of paper and pencil and work through this slowly and carefully.

We have an object being dropped out of an airplane of mass $$m$$. The wind resistance is proportional to the square of the velocity. We want to determine the differential equation associated with this motion and solve for the velocity and position functions. The basic differential equation $$m\dot{v} - m \mu v^2 = -mg$$ is set up in the previous panel. In this section, we show that the solution is

 $$\displaystyle{ v(t) = -V_T \tanh(\omega t) = V_T \left[ \frac{1-e^{2\omega t}}{1+e^{2\omega t}} \right] }$$ where $$\omega = \sqrt{\mu g}$$ and $$V_T = \sqrt{g/\mu}$$

First, we determine where we are going. We start with $$m\dot{v} = m\mu v^2 - mg$$ and we divide through by $$m$$, which we can do since $$m \neq 0$$, to get $$\dot{v} = \mu v^2 - g$$. We are going to use substitution to get $$g\dot{u} = g u^2 - g ~~~ \to ~~~ \dot{u} = u^2 - 1$$.    Notice that, even though this is still nonlinear, it is separable, so we should be able to solve it.

This technique requires us to introduce a variable $$u$$ which is a function of $$s$$. So we have a function $$u(s)$$ and we define $$\displaystyle{ \dot{u} = \frac{du}{ds} }$$.

So now we compare the two equations, term-by-term.
$$\begin{array}{ccccl} \displaystyle{\frac{dv}{dt}} & = & \mu v^2 & - & g \\ \displaystyle{g\frac{du}{ds}} & = & g u^2 & - & g \\ \downarrow & & \downarrow \\ \displaystyle{\frac{dv}{dt} = g\frac{du}{ds}} & & \mu v^2 = g u^2 & & ~~~~~ [ 1 ] \end{array}$$
If we can find a function $$u(s)$$, where $$s$$ is a function of $$t$$, that satisfies these last two equations, we can solve the original differential equations.

Let's start with $$\displaystyle{ \mu v^2 = g u^2 ~~~ \to ~~~ v^2 = \frac{g}{\mu}u^2 ~~~ \to ~~~ v = \sqrt{g/\mu} ~u }$$.
We will call the constant $$\displaystyle{ V_T = \sqrt{g/\mu} }$$ [ terminal velocity - you will see why later ].
So $$v(t) = V_T u(s)$$ where $$V_T$$ is a constant. I wrote the functions with their variables to emphasize each variable.
Let's take the derivative of the last equation with respect to $$t$$.
$$\displaystyle{ \frac{dv}{dt} = V_T \frac{du}{dt}}$$
Since $$u$$ is a function of $$s$$, we use the chain rule to get $$\displaystyle{ \frac{du}{dt} = \frac{du}{ds} \frac{ds}{dt} }$$.

So now we have $$\displaystyle{ \frac{dv}{dt} = V_T \frac{du}{ds} \frac{ds}{dt} ~~~ [ 2 ] }$$

Now let's go back and use equation [1] $$\displaystyle{ \frac{dv}{dt} = g \frac{du}{ds} }$$ and combine it with equation [2] to get
$$\displaystyle{g \frac{du}{ds} = V_T \frac{du}{ds} \frac{ds}{dt} }$$
$$\displaystyle{ g = V_T \frac{ds}{dt} }$$
$$\displaystyle{ \frac{ds}{dt} = \frac{g}{V_T}}$$

Since $$\displaystyle{ V_T = \sqrt{g/\mu} ~~~ \to \frac{ds}{dt} = \frac{g}{\sqrt{g/\mu}} = \sqrt{\mu g}}$$.
We will call this constant $$\omega = \sqrt{\mu g}$$.

Now we have $$\displaystyle{ \frac{ds}{dt} = \omega ~~~ \to ~~~ ds = \omega dt ~~~ \to ~~~ s = \omega t + k }$$; $$k = 0$$ due to the initial conditions.
So we have derived $$s = \omega t$$ where $$\omega = \sqrt{\mu g}$$.

So, finally, we have derived the equations we need to convert $$\dot{v} = \mu v^2 - g$$ to $$\dot{u} = u^2 - 1$$. Let's work forward to check our work.

We will start with $$\displaystyle{ \frac{dv}{dt} = \mu v^2 - g }$$ and let $$v(t) = V_T u(s)$$ where $$\displaystyle{ V_T = \sqrt{g/\mu} }$$.
$$\displaystyle{ v(t) = V_T u(s) }$$
$$\displaystyle{ \frac{dv}{dt} = V_T \frac{du}{dt} }$$
$$\displaystyle{ \frac{dv}{dt} = V_T \frac{du}{ds} \frac{ds}{dt}}$$

$$\displaystyle{ s = \omega t ~~~ \to ~~~ \frac{ds}{dt} = \omega }$$
Now we will combine $$\displaystyle{ \frac{dv}{dt} = \mu v^2 - g }$$ with $$\displaystyle{ \frac{dv}{dt} = V_T \frac{du}{ds} \frac{ds}{dt} }$$ to get

$$\displaystyle{ V_T \omega \frac{du}{ds} = \mu V_T^2 u^2 - g }$$

Next, we will replace $$V_T = \sqrt{g/\mu}$$ and $$\omega = \sqrt{\mu g}$$
$$\displaystyle{ (\sqrt{g/\mu})(\sqrt{\mu g}) \frac{du}{ds} = \mu (\sqrt{g/\mu})^2 u^2 - g }$$
$$\displaystyle{ g \frac{du}{ds} = gu^2 - g }$$
$$\displaystyle{ \frac{du}{ds} = u^2 - 1 }$$
From the initial conditions, we know that $$v(0) = 0$$ since $$s=\omega t$$ so $$s=0$$ when $$t=0$$.
Since $$v(t) = V_T u(s) ~~~ \to ~~~ v(0) = V_T u(0) ~~~ \to ~~~ u(0) = 0$$ when $$s=0$$.
And this is what we wanted.

Recap
Let's recap what we just did and where we go from here.
What we've done so far is show how we can use the equations $$v(t) = V_T u(s); ~~~ V_T = \sqrt{g/\mu} ~~~ \omega = \sqrt{\mu g}$$ to convert $$\displaystyle{ \frac{dv}{dt} = \mu v^2 - g }$$ to $$\displaystyle{ \frac{du}{ds} = u^2 - 1 }$$. The explanation shows where the (strange) constants $$V_T$$ and $$\omega$$ come from.

Now our goal is to solve $$\displaystyle{ \frac{du}{ds} = u^2 - 1 }$$ and use the result to get $$v(t) = V_T u(s)$$ remembering that $$s=\omega t$$. So, let's separate the differential equation and solve.

$$\displaystyle{ \frac{du}{ds} = u^2 - 1 ~~~ \to ~~~ \frac{du}{u^2-1} = ds ~~~ \to ~~~ \int{\frac{du}{u^2-1}} = \int{ds} = s + c_1 }$$

The integral $$\displaystyle{ \int{\frac{du}{u^2-1}} }$$ is solved on the integration by partial fractions page. The result is

$$\displaystyle{ \int{\frac{du}{u^2-1}} = \frac{1}{2}\ln \abs{ \frac{u-1}{u+1} } + c_2 }$$

So we have $$\displaystyle{ \frac{1}{2}\ln \abs{ \frac{u-1}{u+1} } = s + c_3 }$$ where $$c_3 = c_1 - c_2$$.

Now we may be tempted here to solve for $$c_3$$ but the absolute values could give us the wrong answer. So do one more step.

$$\displaystyle{ \frac{u-1}{u+1} = (e^{s+c_3})^2 = Ae^{2s} }$$ where $$A = e^{2c_3}$$.

Now we need to use the initial conditions to determine $$A$$.
$$\displaystyle{ u(0) = 0 ~~~ \to ~~~ \frac{0-1}{0+1} = Ae^0 ~~~ \to ~~~ A = -1 }$$

So now we have $$\displaystyle{ \frac{u-1}{u+1} = -e^{2s} }$$ and we need to solve for $$u$$.

$$\begin{array}{rcl} u - 1 & = & -e^{2s}(u+1) \\ & = & -e^{2s}u-e^{2s} \\ u + ue^{2s} & = & 1 - e^{2s} \\ u(1+e^{2s}) & = & \\ u & = & \displaystyle{\frac{1-e^{2s}}{1+e^{2s}}} \end{array}$$

Note: If you are familiar with hyperbolic trig functions, you may have noticed that
$$\displaystyle{ \int{\frac{du}{u^2-1}} = -\tanh^{-1}(u) }$$ and so $$\displaystyle{ u = -\tanh(s) = \frac{1-e^{2s}}{1+e^{2s}} }$$.

Since $$s = \omega t$$, $$\displaystyle{ u(s) = u(\omega t) = -\tanh(\omega t) = \frac{1-e^{2\omega t}}{1+e^{2\omega t}} }$$.

Substituting this result into $$v(t) = V_T u(s)$$, we finally obtain our result

 $$\displaystyle{ v(t) = -V_T \tanh(\omega t) = V_T \left[ \frac{1-e^{2\omega t}}{1+e^{2\omega t}} \right] }$$ where $$\omega = \sqrt{\mu g}$$ and $$V_T = \sqrt{g/\mu}$$

From the solution in the panel, we have a couple of special constants, $$\omega = \sqrt{\mu g}$$ and $$V_T = \sqrt{g/\mu}$$. So we start with the result of that panel where the velocity function is

$$\displaystyle{v(t) = -V_T ~ \tanh(\omega t) = V_T \left[ \frac{1-e^{2\omega t}}{1+e^{2\omega t}} \right]}$$

Velocity As $$t \to \infty$$

Let's look at what happens to the velocity as $$t$$ approaches infinity by taking the limit

$$\displaystyle{ \lim_{t \to \infty}{V_T \left[ \frac{1-e^{2\omega t}}{1+e^{2\omega t}} \right] } = }$$ $$\displaystyle{ \lim_{t \to \infty}{V_T \left[ \frac{-2\omega e^{2\omega t}}{2\omega e^{2\omega t}} \right] } = -V_T }$$
[We used L'Hôpital's Rule here.]

So as $$t \to \infty$$ the velocity $$v(t) \to -V_T = -\sqrt{g/\mu}$$, which is called the terminal velocity.

Position Function Near $$t=0$$

We need to integrate the velocity function to obtain the position function, i.e. we need to solve $$\displaystyle{ \frac{dy}{dt} = v(t) }$$ and $$y(0) = h_0$$ is the initial height.

$$\displaystyle{ y(t) = \int{v(t)~dt} = \int{-V_t \tanh(\omega t)~dt} }$$

The details for solving the integral $$\displaystyle{ \int{\tanh(\omega t)~dt} }$$ can be found as a practice problem on the integrals page.

So we have $$\displaystyle{ y(t) = \frac{-V_T}{\omega}\ln\abs{\cosh(\omega t)} + k_1 }$$. Using the initial condition, we end up with $$\displaystyle{ y(t) = \frac{-V_T}{\omega}\ln\abs{\cosh(\omega t)} + h_0 }$$

In order to determine what is going on very near $$t=0$$ we replace $$\displaystyle{ \frac{V_T}{\omega}\ln\abs{\cosh(\omega t)} }$$ with it's second degree Taylor series polynomial. The details can be found as a practice problem on the Taylor Series page with the result

$$\displaystyle{ \frac{V_T}{\omega}\ln\abs{\cosh(\omega t)} \approx \frac{\omega^2 t^2}{2\mu} }$$

This gives us $$\displaystyle{ y(t) \approx h_0 - \frac{\omega^2 t^2}{2\mu} = }$$ $$\displaystyle{ h_0 - \frac{\mu g t^2}{2\mu} = }$$ $$\displaystyle{ h_0 - \frac{1}{2}gt^2}$$ for small t.

Summary of Final Answers

differential equation

$$m\dot{v} - m \mu v^2 = -mg$$

velocity function

$$\displaystyle{ v(t) = -V_T ~ \tanh(\omega t) = }$$ $$\displaystyle{ V_T \left[ \frac{1-e^{2\omega t}}{1+e^{2\omega t}} \right] }$$

velocity as $$t \to \infty$$

$$\displaystyle{ \lim_{t \to \infty}{v(t)} = -V_T }$$

position function

$$\displaystyle{ y(t) = \frac{-V_T}{\omega}\ln\abs{\cosh(\omega t)} + h_0 }$$

position function near $$t=0$$

$$\displaystyle{ y(t) \approx h_0 - \frac{1}{2}gt^2}$$

In the process of this solution, we defined the following intermediate values.

 $$\omega = \sqrt{\mu g}$$ $$V_T = \sqrt{g/\mu}$$

Work this practice problem based on this discussion.

A man jumps from an airplane from a height of 1000 feet. If the force due to air resistance is proportional to the square of the velocity (with proportionality constant $$\mu$$), determine his velocity as a function of time. $$\left[ g = 32~ft/sec^2; \mu = 0.0008~ft^{-1} \right]$$

Problem Statement

A man jumps from an airplane from a height of 1000 feet. If the force due to air resistance is proportional to the square of the velocity (with proportionality constant $$\mu$$), determine his velocity as a function of time. $$\left[ g = 32~ft/sec^2; \mu = 0.0008~ft^{-1} \right]$$

$$v(t) = -200\tanh(0.16t)$$

Problem Statement

A man jumps from an airplane from a height of 1000 feet. If the force due to air resistance is proportional to the square of the velocity (with proportionality constant $$\mu$$), determine his velocity as a function of time. $$\left[ g = 32~ft/sec^2; \mu = 0.0008~ft^{-1} \right]$$

Solution

We use the result $$v(t) = -V_T \tanh(\omega t )$$ where $$\omega = \sqrt{\mu g}$$ and $$V_T = \sqrt{g/\mu}$$.
Plugging in the given numbers we have
$$\omega = \sqrt{\mu g} = \sqrt{32(0.0008)} = 4/25 = 0.16$$
$$V_T = \sqrt{g/\mu} = \sqrt{32/0.0008} = 200$$
$$v(t) = -200\tanh(0.16t)$$

$$v(t) = -200\tanh(0.16t)$$