You CAN Ace Differential Equations
external links you may find helpful |
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Single Variable Calculus |
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Multi-Variable Calculus |
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Acceleration Vector |
Arc Length (Vector Functions) |
Arc Length Function |
Arc Length Parameter |
Conservative Vector Fields |
Cross Product |
Curl |
Curvature |
Cylindrical Coordinates |
Lagrange Multipliers |
Line Integrals |
Partial Derivatives |
Partial Integrals |
Path Integrals |
Potential Functions |
Principal Unit Normal Vector |
Differential Equations |
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Do you have a practice problem number but do not know on which page it is found? If so, enter the number below and click 'page' to go to the page on which it is found or click 'practice' to be taken to the practice problem.
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Help Keep 17Calculus Free |
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Welcome to Differential Equations at 17Calculus. Differential Equations is a vast and incredibly fascinating topic that uses calculus extensively. This page gets you started on Ordinary Differential Equations usually covered in a first semester differential equations course.
What Are Differential Equations? |
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Differential Equations consists of a group of techniques used to solve equations that contain derivatives. That's it. That's all there is to it. The complexity comes in because you can't just integrate the equation to solve it. First, you need to classify what kind of differential equation it is based on several criteria. Then, you can choose a technique to solve. Learning to solve differential equations involves learning to classify the equation you are given and then learning the technique to solve that specific type of equation. There is generally no one technique that works in all cases. So, to prepare yourself, spend some extra effort learning to classify the kind of equation you have as you learn each technique. If you don't, you will be totally lost.
There are a lot of shortcuts to solving differential equations. Many instructors teach those shortcuts upfront precisely because they are easier to teach. However, don't let yourself lose sight of where those shortcuts come from and under what conditions you can use them. Spend some time learning the basic technique before using the shortcuts. This usually involves working the first few practice problems with the basic technique. Of course, the instructions your teacher gives you take priority. But really learn these techniques, so that you will know the proper time and situation to use them. After all, that's the point, right? To be able to use this material in your job or other courses?
Notation |
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Constants and Variables |
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One of the first things you need to get your head around with differential equations is which symbols are constants and which are variables. When you see derivative notation you will mostly see \(y'\) instead of \( dy/dt \), for example. So you need to keep track of which symbols are functions, which are variables, what you are taking the derivative with respect to and what are constants.
For example, one equation I ran across in the first section of a differential equations textbook was
\(\displaystyle{ m\frac{dv}{dt} = mg-\gamma v }\)
This could have been written \(\displaystyle{ mv' = mg-\gamma v }\)
In this case, the variable is t and the function is \(v(t)\). The symbols \(m, g, \gamma\) are constants. The context of the problem is important to read and understand in order to arrive at these conclusions.
exp Notation |
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A second thing you need to be aware of is that some textbooks (most of the ones I've seen) use unusual notation for exponential functions. Correct notation is \( y = e^x \). However, sometimes the exponent can be very long and contain a lot of detail. So, the exponential function will sometimes be written as \( y = \exp(x) \). This is only used when the exponent x is detailed. For example,
\(\displaystyle{ \mu(t) = e^{ \int_{t_0}^{t}{p(s)~ds} } }\) is difficult to read. Since there is so much detail in the exponent of \(e\) that we need to see, we usually write this
\(\displaystyle{ \mu(t) = \exp\left( \int_{t_0}^{t}{p(s)~ds} \right) }\)
See how much easier it is to read the exponent? We will follow this convention on this site.
Notation For Derivatives |
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By now you should be comfortable with the notation \(dy/dx\) and \(y'\) for the first derivative. There are a couple of other types of notation that you may or may not have seen before, that you will probably run across on this site, in your textbook, in class and in videos.
\(D_x(y)\) where \(D\) tells you take the derivative and the subscript x is the variable of integration. |
\(\dot{y}\) where the dot above the y tells you to take the first derivative (two dots for the second derivative). |
Classifying Differential Equations |
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One key to success in solving differential equations involves classifying the type of differential equation you have. Correct identification will narrow down the list of available techniques that you need to choose from. Specific techniques work only with specific types of differential equations. So you need to start by looking carefully at the type of equation you have. Here is a broad list of types.
Order |
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Order indicates the highest derivative appearing in the differential equation. For example, \( y' - y = 0 ~\) is a first order differential equation because the highest derivative is a first derivative. Similarly, \( y'' - y = 0 \) is a second order differential equation because the highest derivative is a second derivative.
Linear |
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A differential equation of the form \(P(t) y'' + Q(t)y' + R(t)y = G(t) \) is linear since \(P(t)\), \(Q(t)\) and \(R(t)\) are functions of t only, i.e. they do not contain any y's or derivatives of y. Analyzing nonlinear equations is relatively difficult, so it is unlikely you will encounter them in a first semester differential equations course, except under very special circumstances.
video by MIP4U
Homogeneous |
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There are several meanings of the term 'homogeneous' used in differential equations. For this definition, assume you have a differential equation of the form \( p(t)y'' + q(t)y' + r(t)y = g(t) \).
Meaning 1 - - If \( g(t) = 0 \) in the above form, then the differential equation is said to be homogeneous. The idea is to get all the terms containing y or a derivative of y to one side of the equal sign and all other terms to the other side. If there are no terms without a y or it's derivatives, then \(g(t)\) will be zero and the equation will be homogeneous. If \(g(t) \neq 0 \), then the differential equation is said to be inhomogeneous (or nonhomogeneous).
Meaning 2 - - If all of the expressions \(p(t)\), \(q(t)\), \(r(t)\) and \(g(t)\) can be written as functions of \(y/t\), then it is said to be homogeneous. In this case, we use the technique of substitution to solve this type of differential equation.
Before we go on, let's do a few practice problems related to classifying differential equations.
Instructions - Classify these differential equations by order and type.
\(\displaystyle{ y \frac{d^2y}{dx^2} + 4 \frac{dy}{dx} = \sin(x) }\)
Problem Statement |
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Classify the differential equation \(\displaystyle{ y \frac{d^2y}{dx^2} + 4 \frac{dy}{dx} = \sin(x) }\) by order and type.
Final Answer |
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second order, nonlinear, nonhomogeneous |
Problem Statement |
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Classify the differential equation \(\displaystyle{ y \frac{d^2y}{dx^2} + 4 \frac{dy}{dx} = \sin(x) }\) by order and type.
Solution |
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The differential equation is
- second order due to the \(\displaystyle{ \frac{d^2y}{dx^2} }\) term
- nonlinear due to multiplication of y in \(\displaystyle{ \left[ y\frac{d^2y}{dx^2} \right] }\)
- nonhomogeneous due to the \( \sin(x) \) term.
Final Answer |
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second order, nonlinear, nonhomogeneous |
close solution |
\(\displaystyle{ \frac{d^3 y}{dy^{3}} + 4\sin(x)~y = 0 }\)
Problem Statement |
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Classify the differential equation \(\displaystyle{ \frac{d^3 y}{dy^{3}} + 4\sin(x)~y = 0 }\) by order and type.
Final Answer |
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third order, linear, homogeneous |
Problem Statement |
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Classify the differential equation \(\displaystyle{ \frac{d^3 y}{dy^{3}} + 4\sin(x)~y = 0 }\) by order and type.
Solution |
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The differential equation is
- third order due to the \(\displaystyle{ \frac{d^3 y}{dy^3} }\) term
- linear
- homogeneous since the right side is equal to zero
Final Answer |
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third order, linear, homogeneous |
close solution |
General vs Particular Solution |
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What is the difference between the general solution and the particular solution?
Short Answer - - The terminology relates to whether or not we have an unknown constant in our final answer. For the general solution, an unknown constant IS part of the solution indicating an infinite number of solutions. In the particular solution, there is no unknown constant and the solution we have is the one and only solution.
Long Answer - - When you solve a differential equation, you use integration, which introduces an unknown constant.
In the general solution, the unknown constant remains and you do not have enough information to be able to determine what that constant is. Consequently, you have an infinite number of solutions.
In the particular solution, you start out by solving for the general solution but then you are given initial conditions which you use to determine the value of the constant. These initial conditions are actually points that the solution to the differential equation pass through. In the end, you have only one solution without any unknown constants.
general solution |
particular solution | |
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infinite number of solutions |
only one solution | |
contains unknown constant(s) |
does NOT contain any unknown constants | |
no (or too few) initial conditions given |
initial conditions given; used to solve for constants |
To find the general solution, just solve the differential equation and leave any unknown constants in your final answer.
To find the particular solution, find the general solution first, then plug in the initial conditions and solve for the constants.
Notes
1. When determining the particular solution, you will be given the same number of initial conditions as the order of the differential equation. Depending on how the initial conditions are given, you may need to stop after each integration and solve for the individual constants or you may need to wait until you are completely done and solve for all the constants at once. You will get a feel for this as you work practice problems.
2. For a general solution, if no initial conditions are given or fewer than the order of the differential equation are given, we cannot determine all of the unknown constants, since each integration introduces another constant.
3. When you were first learning integration, you probably ran across initial value problems. These were actually differential equations where you were asked to find the particular solution.
Getting Started |
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After going through the above information you are ready to watch some videos to get started with differential equations.
Here is a good introduction to differential equations. He contrasts a differential equation to a standard equation, which you should be familiar with, and explains, practically, what a differential equation is. He also works the example \( y'' + 2y' - 3y = 0 \) and shows that \( y_1 = e^{-3x} \) and \( y_2 = e^x \) are solutions to this differential equation. Then, he goes on to explain linear versus nonlinear and order.
video by Khan Academy
Here is another introduction video. The technique he uses is separation of variables, which is the first technique usually introduced in a differential equations course. It will help you to see this technique in the context of introducing differential equations.
video by Khan Academy
Here is a good video showing what it means for an equation to be a solution to a differential equation. This also demonstrates how to check your answer after you have solved a differential equation.
video by PatrickJMT
Okay, let's work a few practice problems before going on.
Show that \( y(x) = \tanh(x+a) \) satisfies \( y'=1-y^2 \).
Problem Statement |
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Show that \( y(x) = \tanh(x+a) \) satisfies \( y'=1-y^2 \).
Hint |
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To determine if a function satisfies a differential equation, plug it into the original differential equation and check that the equation holds.
Problem Statement |
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Show that \( y(x) = \tanh(x+a) \) satisfies \( y'=1-y^2 \).
Hint |
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To determine if a function satisfies a differential equation, plug it into the original differential equation and check that the equation holds.
Solution |
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\( y = \tanh(x+a) \) \( \to \) \( y' = \sech^2(x+a) \)
Let's rearrange the differential equation
\( y'=1-y^2 \) \( \to \) \( y' + y^2 -1 = 0 \)
\( y' + y^2 - 1 \) |
\( \sech^2(x+a) + \tanh^2(x+a) - 1 \) |
\( \displaystyle{\frac{1}{\cosh^2(x+a)} + \frac{\sinh^2(x+a)}{\cosh^2(x+a)} - \frac{\cosh^2(x+a)}{\cosh^2(x+a)}} \) |
\( \displaystyle{\frac{1+\sinh^2(x+a)-\cosh^2(x+a)}{\cosh^2(x+a)}} \) |
Use the hyperbolic identity \( \cosh^2(x+a) - \sinh^2(x+a) = 1 \) |
\( \displaystyle{\frac{1-1}{\cosh^2(x+a)} = 0} ~~~ \text{[qed]} \) |
close solution |
Now that you have an overview of differential equations, you are ready to begin studying specific topics. The next natural topic is slope fields. However, many instructors will go straight to separation of variables. In any case, take some time to enjoy studying differential equations.