- Last updated
- Save as PDF
- Page ID
- 10782
\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}}}\) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)\(\newcommand{\AA}{\unicode[.8,0]{x212B}}\)
Learning Objectives
- Identify the order of a differential equation.
- Explain what is meant by a solution to a differential equation.
- Distinguish between the general solution and a particular solution of a differential equation.
- Identify an initial-value problem.
- Identify whether a given function is a solution to a differential equation or an initial-value problem.
Calculus is the mathematics of change, and rates of change are expressed by derivatives. Thus, one of the most common ways to use calculus is to set up an equation containing an unknown function \(y=f(x)\) and its derivative, known as a differential equation. Solving such equations often provides information about how quantities change and frequently provides insight into how and why the changes occur.
Techniques for solving differential equations can take many different forms, including direct solution, use of graphs, or computer calculations. We introduce the main ideas in this chapter and describe them in a little more detail later in the course. In this section we study what differential equations are, how to verify their solutions, some methods that are used for solving them, and some examples of common and useful equations.
General Differential Equations
Consider the equation \(y′=3x^2,\) which is an example of a differential equation because it includes a derivative. There is a relationship between the variables \(x\) and \(y:y\) is an unknown function of \(x\). Furthermore, the left-hand side of the equation is the derivative of \(y\). Therefore we can interpret this equation as follows: Start with some function \(y=f(x)\) and take its derivative. The answer must be equal to \(3x^2\). What function has a derivative that is equal to \(3x^2\)? One such function is \(y=x^3\), so this function is considered a solution to a differential equation.
Definition: differential equation
A differential equation is an equation involving an unknown function \(y=f(x)\) and one or more of its derivatives. A solution to a differential equation is a function \(y=f(x)\) that satisfies the differential equation when \(f\) and its derivatives are substituted into the equation.
Go to this website to explore more on this topic.
Some examples of differential equations and their solutions appear in Table \(\PageIndex{1}\).
| Equation | Solution |
|---|---|
| \(y'=2x\) | \(y=x^2\) |
| \(y'+3y=6x+11\) | \(y=e^{−3x}+2x+3\) |
| \(y''−3y'+2y=24e^{−2x}\) | \(y=3e^x−4e^{2x}+2e^{−2x}\) |
Note that a solution to a differential equation is not necessarily unique, primarily because the derivative of a constant is zero. For example, \(y=x^2+4\) is also a solution to the first differential equation in Table \(\PageIndex{1}\). We will return to this idea a little bit later in this section. For now, let’s focus on what it means for a function to be a solution to a differential equation.
Example \(\PageIndex{1}\): Verifying Solutions of Differential Equations
Verify that the function \(y=e^{−3x}+2x+3\) is a solution to the differential equation \(y′+3y=6x+11\).
Solution
To verify the solution, we first calculate \(y′\) using the chain rule for derivatives. This gives \(y′=−3e^{−3x}+2\). Next we substitute \(y\) and \(y′\) into the left-hand side of the differential equation:
\((−3e^{−2x}+2)+3(e^{−2x}+2x+3).\)
The resulting expression can be simplified by first distributing to eliminate the parentheses, giving
\(−3e^{−2x}+2+3e^{−2x}+6x+9.\)
Combining like terms leads to the expression \(6x+11\), which is equal to the right-hand side of the differential equation. This result verifies that \(y=e^{−3x}+2x+3\) is a solution of the differential equation.
Exercise \(\PageIndex{1}\)
Verify that \(y=2e^{3x}−2x−2\) is a solution to the differential equation \(y′−3y=6x+4.\)
- Hint
-
First calculate \(y′\) then substitute both \(y′\) and \(y\) into the left-hand side.
It is convenient to define characteristics of differential equations that make it easier to talk about them and categorize them. The most basic characteristic of a differential equation is its order.
Definition: order of a differential equation
The order of a differential equation is the highest order of any derivative of the unknown function that appears in the equation.
Example \(\PageIndex{2}\): Identifying the Order of a Differential Equation
The highest derivative in the equation is \(y′\),
What is the order of each of the following differential equations?
- \(y′−4y=x^2−3x+4\)
- \(x^2y'''−3xy''+xy′−3y=\sin x\)
- \(\frac{4}{x}y^{(4)}−\frac{6}{x^2}y''+\frac{12}{x^4}y=x^3−3x^2+4x−12\)
Solution
- The highest derivative in the equation is \(y′\),so the order is \(1\).
- The highest derivative in the equation is \(y'''\), so the order is \(3\).
- The highest derivative in the equation is \(y^{(4)}\), so the order is \(4\).
Exercise \(\PageIndex{2}\)
What is the order of the following differential equation?
\((x^4−3x)y^{(5)}−(3x^2+1)y′+3y=\sin x\cos x\)
- Hint
-
What is the highest derivative in the equation?
- Answer
-
\(5\)
General and Particular Solutions
We already noted that the differential equation \(y′=2x\) has at least two solutions: \(y=x^2\) and \(y=x^2+4\). The only difference between these two solutions is the last term, which is a constant. What if the last term is a different constant? Will this expression still be a solution to the differential equation? In fact, any function of the form \(y=x^2+C\), where \(C\) represents any constant, is a solution as well. The reason is that the derivative of \(x^2+C\) is \(2x\), regardless of the value of \(C\). It can be shown that any solution of this differential equation must be of the form \(y=x^2+C\). This is an example of a general solution to a differential equation. A graph of some of these solutions is given in Figure \(\PageIndex{1}\). (Note: in this graph we used even integer values for C ranging between \(−4\) and \(4\). In fact, there is no restriction on the value of \(C\); it can be an integer or not.)
In this example, we are free to choose any solution we wish; for example, \(y=x^2−3\) is a member of the family of solutions to this differential equation. This is called a particular solution to the differential equation. A particular solution can often be uniquely identified if we are given additional information about the problem.
Example \(\PageIndex{3}\): Finding a Particular Solution
Find the particular solution to the differential equation \(y′=2x\) passing through the point \((2,7)\).
Solution
Any function of the form \(y=x^2+C\) is a solution to this differential equation. To determine the value of \(C\), we substitute the values \(x=2\) and \(y=7\) into this equation and solve for \(C\):
\[ \begin{align*} y =x^2+C \\[4pt] 7 =2^2+C \\[4pt] =4+C \\[4pt] C =3. \end{align*}\]
Therefore the particular solution passing through the point \((2,7)\) is \(y=x^2+3\).
Exercise \(\PageIndex{3}\)
Find the particular solution to the differential equation
\[ y′=4x+3 \nonumber \]
passing through the point \((1,7),\) given that \(y=2x^2+3x+C\) is a general solution to the differential equation.
- Hint
-
First substitute \(x=1\) and \(y=7\) into the equation, then solve for \(C\).
- Answer
-
\[ y=2x^2+3x+2 \nonumber \]
Initial-Value Problems
Usually a given differential equation has an infinite number of solutions, so it is natural to ask which one we want to use. To choose one solution, more information is needed. Some specific information that can be useful is an initial value, which is an ordered pair that is used to find a particular solution.
A differential equation together with one or more initial values is called an initial-value problem. The general rule is that the number of initial values needed for an initial-value problem is equal to the order of the differential equation. For example, if we have the differential equation \(y′=2x\), then \(y(3)=7\) is an initial value, and when taken together, these equations form an initial-value problem. The differential equation \(y''−3y′+2y=4e^x\) is second order, so we need two initial values. With initial-value problems of order greater than one, the same value should be used for the independent variable. An example of initial values for this second-order equation would be \(y(0)=2\) and \(y′(0)=−1.\) These two initial values together with the differential equation form an initial-value problem. These problems are so named because often the independent variable in the unknown function is \(t\), which represents time. Thus, a value of \(t=0\) represents the beginning of the problem.
Example \(\PageIndex{4}\): Verifying a Solution to an Initial-Value Problem
Verify that the function \(y=2e^{−2t}+e^t\) is a solution to the initial-value problem
\[ y′+2y=3e^t, \quad y(0)=3.\nonumber \]
Solution
For a function to satisfy an initial-value problem, it must satisfy both the differential equation and the initial condition. To show that \(y\) satisfies the differential equation, we start by calculating \(y′\). This gives \(y′=−4e^{−2t}+e^t\). Next we substitute both \(y\) and \(y′\) into the left-hand side of the differential equation and simplify:
\[ \begin{align*} y′+2y &=(−4e^{−2t}+e^t)+2(2e^{−2t}+e^t) \\[4pt] &=−4e^{−2t}+e^t+4e^{−2t}+2e^t =3e^t. \end{align*}\]
This is equal to the right-hand side of the differential equation, so \(y=2e^{−2t}+e^t\) solves the differential equation. Next we calculate \(y(0)\):
\[ y(0)=2e^{−2(0)}+e^0=2+1=3. \nonumber \]
This result verifies the initial value. Therefore the given function satisfies the initial-value problem.
Exercise \(\PageIndex{4}\)
Verify that \(y=3e^{2t}+4\sin t\) is a solution to the initial-value problem
\[ y′−2y=4\cos t−8\sin t,y(0)=3. \nonumber \]
- Hint
-
First verify that \(y\) solves the differential equation. Then check the initial value.
In Example \(\PageIndex{4}\), the initial-value problem consisted of two parts. The first part was the differential equation \(y′+2y=3e^x\), and the second part was the initial value \(y(0)=3.\) These two equations together formed the initial-value problem.
The same is true in general. An initial-value problem will consists of two parts: the differential equation and the initial condition. The differential equation has a family of solutions, and the initial condition determines the value of \(C\). The family of solutions to the differential equation in Example \(\PageIndex{4}\) is given by \(y=2e^{−2t}+Ce^t.\) This family of solutions is shown in Figure \(\PageIndex{2}\), with the particular solution \(y=2e^{−2t}+e^t\) labeled.
Example \(\PageIndex{5}\): Solving an Initial-value Problem
Solve the following initial-value problem:
\[ y′=3e^x+x^2−4,y(0)=5. \nonumber \]
Solution
The first step in solving this initial-value problem is to find a general family of solutions. To do this, we find an antiderivative of both sides of the differential equation
\[∫y′\,dx=∫(3e^x+x^2−4)\,dx, \nonumber \]
namely,
\(y+C_1=3e^x+\frac{1}{3}x^3−4x+C_2\).
We are able to integrate both sides because the y term appears by itself. Notice that there are two integration constants: \(C_1\) and \(C_2\). Solving this equation for \(y\) gives
\(y=3e^x+\frac{1}{3}x^3−4x+C_2−C_1.\)
Because \(C_1\) and \(C_2\) are both constants, \(C_2−C_1\) is also a constant. We can therefore define \(C=C_2−C_1,\) which leads to the equation
\(y=3e^x+\frac{1}{3}x^3−4x+C.\)
Next we determine the value of \(C\). To do this, we substitute \(x=0\) and \(y=5\) into this equation and solve for \(C\):
\[ \begin{align*} 5 &=3e^0+\frac{1}{3}0^3−4(0)+C \\[4pt] 5 &=3+C \\[4pt] C&=2 \end{align*}. \nonumber \]
Now we substitute the value \(C=2\) into the general equation. The solution to the initial-value problem is \(y=3e^x+\frac{1}{3}x^3−4x+2.\)
Analysis
The difference between a general solution and a particular solution is that a general solution involves a family of functions, either explicitly or implicitly defined, of the independent variable. The initial value or values determine which particular solution in the family of solutions satisfies the desired conditions.
Exercise \(\PageIndex{5}\)
Solve the initial-value problem
\[ y′=x^2−4x+3−6e^x,y(0)=8. \nonumber \]
- Hint
-
First take the antiderivative of both sides of the differential equation. Then substitute \(x=0\) and \(y=8\) into the resulting equation and solve for \(C\).
- Answer
-
\(y=\frac{1}{3}x^3−2x^2+3x−6e^x+14\)
In physics and engineering applications, we often consider the forces acting upon an object, and use this information to understand the resulting motion that may occur. For example, if we start with an object at Earth’s surface, the primary force acting upon that object is gravity. Physicists and engineers can use this information, along with Newton’s second law of motion (in equation form \(F=ma\), where \(F\) represents force, \(m\) represents mass, and \(a\) represents acceleration), to derive an equation that can be solved.
In Figure \(\PageIndex{3}\) we assume that the only force acting on a baseball is the force of gravity. This assumption ignores air resistance. (The force due to air resistance is considered in a later discussion.) The acceleration due to gravity at Earth’s surface, g, is approximately \(9.8\,\text{m/s}^2\). We introduce a frame of reference, where Earth’s surface is at a height of 0 meters. Let \(v(t)\) represent the velocity of the object in meters per second. If \(v(t)>0\), the ball is rising, and if \(v(t)<0\), the ball is falling (Figure).
Our goal is to solve for the velocity \(v(t)\) at any time \(t\). To do this, we set up an initial-value problem. Suppose the mass of the ball is \(m\), where \(m\) is measured in kilograms. We use Newton’s second law, which states that the force acting on an object is equal to its mass times its acceleration \((F=ma)\). Acceleration is the derivative of velocity, so \(a(t)=v′(t)\). Therefore the force acting on the baseball is given by \(F=mv′(t)\). However, this force must be equal to the force of gravity acting on the object, which (again using Newton’s second law) is given by \(F_g=−mg\), since this force acts in a downward direction. Therefore we obtain the equation \(F=F_g\), which becomes \(mv′(t)=−mg\). Dividing both sides of the equation by \(m\) gives the equation
\[ v′(t)=−g. \nonumber \]
Notice that this differential equation remains the same regardless of the mass of the object.
We now need an initial value. Because we are solving for velocity, it makes sense in the context of the problem to assume that we know the initial velocity, or the velocity at time \(t=0.\) This is denoted by \(v(0)=v_0.\)
Example \(\PageIndex{6}\): Velocity of a Moving Baseball
A baseball is thrown upward from a height of \(3\) meters above Earth’s surface with an initial velocity of \(10\) m/s, and the only force acting on it is gravity. The ball has a mass of \(0.15\) kg at Earth’s surface.
- Find the velocity \(v(t)\) of the basevall at time \(t\).
- What is its velocity after \(2\) seconds?
Solution
a. From the preceding discussion, the differential equation that applies in this situation is
\(v′(t)=−g,\)
where \(g=9.8\, \text{m/s}^2\). The initial condition is \(v(0)=v_0\), where \(v_0=10\) m/s. Therefore the initial-value problem is \(v′(t)=−9.8\,\text{m/s}^2,\,v(0)=10\) m/s.
The first step in solving this initial-value problem is to take the antiderivative of both sides of the differential equation. This gives
\[\int v′(t)\,dt=∫−9.8\,dt \nonumber \]
\(v(t)=−9.8t+C.\)
The next step is to solve for \(C\). To do this, substitute \(t=0\) and \(v(0)=10\):
\[ \begin{align*} v(t) &=−9.8t+C \\[4pt] v(0) &=−9.8(0)+C \\[4pt] 10 &=C. \end{align*}\]
Therefore \(C=10\) and the velocity function is given by \(v(t)=−9.8t+10.\)
b. To find the velocity after \(2\) seconds, substitute \(t=2\) into \(v(t)\).
\[ \begin{align*} v(t)&=−9.8t+10 \\[4pt] v(2)&=−9.8(2)+10 \\[4pt] v(2) &=−9.6\end{align*}\]
The units of velocity are meters per second. Since the answer is negative, the object is falling at a speed of \(9.6\) m/s.
Exercise \(\PageIndex{6}\)
Suppose a rock falls from rest from a height of \(100\) meters and the only force acting on it is gravity. Find an equation for the velocity \(v(t)\) as a function of time, measured in meters per second.
- Hint
-
What is the initial velocity of the rock? Use this with the differential equation in Example \(\PageIndex{6}\) to form an initial-value problem, then solve for \(v(t)\).
- Answer
-
\(v(t)=−9.8t\)
A natural question to ask after solving this type of problem is how high the object will be above Earth’s surface at a given point in time. Let \(s(t)\) denote the height above Earth’s surface of the object, measured in meters. Because velocity is the derivative of position (in this case height), this assumption gives the equation \(s′(t)=v(t)\). An initial value is necessary; in this case the initial height of the object works well. Let the initial height be given by the equation \(s(0)=s_0\). Together these assumptions give the initial-value problem
\[ s′(t)=v(t),s(0)=s_0. \nonumber \]
If the velocity function is known, then it is possible to solve for the position function as well.
Example \(\PageIndex{7}\): Height of a Moving Baseball
A baseball is thrown upward from a height of \(3\) meters above Earth’s surface with an initial velocity of \(10m/s\), and the only force acting on it is gravity. The ball has a mass of \(0.15\) kilogram at Earth’s surface.
- Find the position \(s(t)\) of the baseball at time \(t\).
- What is its height after \(2\) seconds?
Solution
We already know the velocity function for this problem is \(v(t)=−9.8t+10\). The initial height of the baseball is \(3\) meters, so \(s_0=3\). Therefore the initial-value problem for this example is
To solve the initial-value problem, we first find the antiderivatives:
\[∫s′(t)\,dt=∫(−9.8t+10)\,dt \nonumber \]
\(s(t)=−4.9t^2+10t+C.\)
Next we substitute \(t=0\) and solve for \(C\):
\(s(t)=−4.9t^2+10t+C\)
\(s(0)=−4.9(0)^2+10(0)+C\)
\(3=C\).
Therefore the position function is \(s(t)=−4.9t^2+10t+3.\)
b. The height of the baseball after \(2\) sec is given by \(s(2):\)
\(s(2)=−4.9(2)^2+10(2)+3=−4.9(4)+23=3.4.\)
Therefore the baseball is \(3.4\) meters above Earth’s surface after \(2\) seconds. It is worth noting that the mass of the ball cancelled out completely in the process of solving the problem.
Key Concepts
- A differential equation is an equation involving a function \(y=f(x)\) and one or more of its derivatives. A solution is a function \(y=f(x)\) that satisfies the differential equation when \(f\) and its derivatives are substituted into the equation.
- The order of a differential equation is the highest order of any derivative of the unknown function that appears in the equation.
- A differential equation coupled with an initial value is called an initial-value problem. To solve an initial-value problem, first find the general solution to the differential equation, then determine the value of the constant. Initial-value problems have many applications in science and engineering.
Glossary
- differential equation
- an equation involving a function \(y=y(x)\) and one or more of its derivatives
- general solution (or family of solutions)
- the entire set of solutions to a given differential equation
- initial value(s)
- a value or set of values that a solution of a differential equation satisfies for a fixed value of the independent variable
- initial velocity
- the velocity at time \(t=0\)
- initial-value problem
- a differential equation together with an initial value or values
- order of a differential equation
- the highest order of any derivative of the unknown function that appears in the equation
- particular solution
- member of a family of solutions to a differential equation that satisfies a particular initial condition
- solution to a differential equation
- a function \(y=f(x)\) that satisfies a given differential equation
FAQs
8.1: Basics of Differential Equations? ›
In general, solving an ODE is more complicated than simple integration. Even so, the basic principle is always integration, as we need to go from derivative to function. Usually, the difficult part is determining what integration we need to do.
Is ordinary differential equations hard? ›In general, solving an ODE is more complicated than simple integration. Even so, the basic principle is always integration, as we need to go from derivative to function. Usually, the difficult part is determining what integration we need to do.
What difficulty level is differential equations? ›An undergraduate differential equations course is easier than calculus, in that there are not actually any new ideas. All the ingredients are directly taken from calculus, whereas calculus includes some topology as well as derivations.
What math do I need to know for differential equations? ›The prerequisites are calculus and linear algebra. No other prerequisites are needed. It's not a very difficult course so it's a good one to take immediately after taking linear algebra.
How do you solve differential equations easily? ›- Substitute y = uv, and. ...
- Factor the parts involving v.
- Put the v term equal to zero (this gives a differential equation in u and x which can be solved in the next step)
- Solve using separation of variables to find u.
- Substitute u back into the equation we got at step 2.
At an advanced level, statistics is considered harder than calculus, but beginner-level statistics is much easier than beginner calculus.
What is the hardest math course in college? ›Advanced Calculus is the hardest math subject, according to college professors. One of the main reasons students struggle to understand the concepts in Advanced Calculus is because they do not have a good mathematical foundation. Calculus builds on the algebraic concepts learned in previous classes.
Is diff eq harder than linear algebra? ›I would say they're roughly equal in difficulty. Differential equations will require more memorization of techniques, but you might find them very intuitive. There will be very few proofs. This is probably because they're too advanced for a first course.
Is differential equations above calculus? ›Two main courses after calculus are linear algebra and differential equations. I hope you can take both. To help you later, Sections 16.1 and 16.2 organize them by examples.
What should I study to prepare for differential equations? ›To confidently solve differential equations, you need to understand how the equations are classified by order, how to distinguish between linear, separable, and exact equations, and how to identify homogenous and nonhomogeneous differential equations.
What should I study before differential equations? ›
To begin, we'll discuss some simple principles that you should understand prior to trying to learn differential equations. These foundations are important, and include basic Calculus, Algebra, and Arithmetic.
Is there a calculator that can solve differential equations? ›To solve ordinary differential equations (ODEs) use the Symbolab calculator.
How fast can you learn differential equations? ›How long does it take to learn differential equations? That depends what you mean. You could probably learn what a differential equation is in a few minutes on Wikipedia. High-school calculus classes usually spend a couple of weeks on simple first-order linear differential equations.
How many methods are there to solve differential equations? ›There exist two methods to find the solution of the differential equation.
Do most students fail calculus? ›Those concerns were heightened given significant declines in recent standardized math test scores for K-12 California students, most of whom spent 2020-21 in distance learning. CSU Bakersfield reports that 35.5% of students on average have been failing or withdrawing from Calculus 1 since 2018.
What is the hardest math? ›- Separatrix Separation. A pendulum in motion can either swing from side to side or turn in a continuous circle. ...
- Navier–Stokes. ...
- Exponents and dimensions. ...
- Impossibility theorems. ...
- Spin glass.
x3+y3+z3=k, with k being all the numbers from one to 100, is a Diophantine equation that's sometimes known as "summing of three cubes." When there are two or more unknowns, as is the case here, only the integers are studied.
Is there really a math 55 at Harvard? ›Math 55 is a two-semester long first-year undergraduate mathematics course at Harvard University, founded by Lynn Loomis and Shlomo Sternberg. The official titles of the course are Honors Abstract Algebra (Math 55a) and Honors Real and Complex Analysis (Math 55b).
Did Bill Gates pass math 55? ›Bill Gates took Math 55.
To get a sense of the kind of brains it takes to get through Math 55, consider that Bill Gates himself was a student in the course. (He passed.) And if you'd like to sharpen your brain like Microsoft's co-founder, here are The 5 Books Bill Gates Says You Should Read.
Similarly, the B-level conventional course students failed Calculus 2 at a rate of 17.6%, while the B-level extended course students had a much lower Calculus 2 failure rate of 10.1%.
Should I do Calc 3 or differential equations first? ›
After completing Calculus I and II, you may continue to Calculus III, Linear Algebra, and Differential Equations. These three may be taken in any order that fits your schedule, but the listed order is most common.
How many people fail linear algebra? ›Linear algebra 1 - about 45% fail rate.
What math is higher than differential equations? ›Appropriate topics to study once you understand calculus and differential equations include: Real analysis, which is a more rigorous approach to calculus of real-valued functions. Complex analysis, which concerns the calculus applied to complex-valued functions of a complex variable.
What grade level is solving linear equations? ›Grade 8: Solving Linear Equations in One Variable.
What is the highest level of math in college? ›A doctoral degree is the highest level of education available in mathematics, often taking 4-7 years to complete. Like a master's degree, these programs offer specializations in many areas, including computer algebra, mathematical theory analysis, and differential geometry.
What is the easiest branch of math? ›GEOMETRY: This is one of the most favorite and easiest branches of mathematics. This branch deals with the shapes and sizes of figures and their properties. Point, line, angle, surface, and solid entities constitute the basic elements of geometry.
What percent of the world knows calculus? ›If 85% of adults graduate high school, and only 16% of those take take calculus, then 13% of adults in the developed world study calculus.
What is the best program to solve differential equations? ›- MATLAB Central.
- ODE Software for MATLAB.
- Books on MATLAB.
The laws of physics are generally written down as differential equations. Therefore, all of science and engineering use differential equations to some degree. Understanding differential equations is essential to understanding almost anything you will study in your science and engineering classes.
What are the 4 types of differential equations? ›While differential equations have three basic types—ordinary (ODEs), partial (PDEs), or differential-algebraic (DAEs), they can be further described by attributes such as order, linearity, and degree.
Can we study differential equations without studying integration? ›
you need to have complete knowledge of both differentiation and integration to understand and solve problems on differential equations.
Is differential equations needed for finance? ›Differential equations are the necessary tool for dealing with problems involving “contin- uously compounded” interest. Particularly in the cases of savings accounts, fixed rate bonds, annuities, mortgages, etc.
Do you need to know linear algebra for differential equations? ›Differential equations are both challenging objects at a mathematical level and crucial in many ways for engineers. In addition, linear algebra methods are an essential part of the methodology commonly used in order to solve systems of differential equations.
Is differential equations just calculus? ›In mathematics, differential calculus is a subfield of calculus that studies the rates at which quantities change. It is one of the two traditional divisions of calculus, the other being integral calculus—the study of the area beneath a curve.
Should I learn differential or integral first? ›Answer and Explanation:
We know that while solving the integration questions, there is a wide application of derivatives. Hence, the differential calculus is taught before the integral calculus. Also, we can evaluate the indefinite integrals easily if we understand the concept of the differential calculus.
You need calculus I and II for sure. A differential equation is an equation involving derivatives, and the first topic usually covered in DE is a form of integration. So you may actually be able to get by with just Calculus I, but I would suggest following the required prerequisites.
Is it necessary to learn differential equations? ›Differential equations are important because for many physical systems, one can, subject to suitable idealizations, formulate a differential equation that describes how the system changes in time. Understanding the solutions of the differential equation is then of paramount interest.
What is the 3 step rule in differential calculus? ›- Substitute your function into the limit definition formula.
- Simplify as needed.
- Evaluate the limit.
It depends on how the course is taught. But generally, it is similar to calculus. It may rely on some linear algebra knowledge, so you may want to make sure that you are not missing that part of the prerequisites.
What level of Calc is differential equations? ›Differential equations are defined in the second semester of calculus as a generalization of antidifferentiation and strategies for addressing the simplest types are addressed there.
What are the hardest types of differential equations? ›
Nonlinear differential equations are often very difficult or impossible to solve. One approach getting around this difficulty is to linearize the differential equation.
What is the most accurate ODE solver? ›ode45 performs well with most ODE problems and should generally be your first choice of solver. However, ode23 , ode78 , ode89 and ode113 can be more efficient than ode45 for problems with looser or tighter accuracy requirements. Some ODE problems exhibit stiffness, or difficulty in evaluation.
Which method is fastest to give solution of differential equations? ›Euler's method is the fastest possible single-step, explicit, non-adaptive method for a given fixed step size h.
Is ordinary differential equations harder than calculus? ›At a basic level, I find that multivariable calculus requires a specific type of spacial thinking that can be very challenging while differential equations is more just about recognizing patterns and types of equations. For many, multivariable calculus will be much more challenging.
Are partial differential equations harder than ordinary? ›An ode contains ordinary derivatives and a pde contains partial derivatives. Typically, pde's are much harder to solve than ode's.
What is the hardest calculus class in college? ›Those who have difficulties memorizing and applying new, unrelated mathematical techniques may say that Calculus 2 is the most challenging calculus class. On the other hand, students who struggle with making three-dimensional calculations may argue that Calculus 3 is the most difficult.
What is the hardest calculus in high school? ›What is the Hardest Math Class in High School? In most cases, you'll find that AP Calculus BC or IB Math HL is the most difficult math course your school offers. Note that AP Calculus BC covers the material in AP Calculus AB but also continues the curriculum, addressing more challenging and advanced concepts.
What math class is higher than calculus? ›After completing Calculus I and II, you may continue to Calculus III, Linear Algebra, and Differential Equations. These three may be taken in any order that fits your schedule, but the listed order is most common.
What is the hardest math linear equation? ›For decades, a math puzzle has stumped the smartest mathematicians in the world. x3+y3+z3=k, with k being all the numbers from one to 100, is a Diophantine equation that's sometimes known as "summing of three cubes."
What is the fail rate for calculus 2? ›Similarly, the B-level conventional course students failed Calculus 2 at a rate of 17.6%, while the B-level extended course students had a much lower Calculus 2 failure rate of 10.1%.
How much calculus is needed for differential equations? ›
You need calculus I and II for sure. A differential equation is an equation involving derivatives, and the first topic usually covered in DE is a form of integration. So you may actually be able to get by with just Calculus I, but I would suggest following the required prerequisites.
Do engineers learn partial differential equations? ›Partial differential equations are ubiquitous in mathematically oriented scientific fields, such as physics and engineering.
Why is PDE so hard? ›The hard part is finding the solutions to a given PDE. Solution spaces tend to be infinite dimensional. It's not usually possible to write down every solution. There are no known techniques that will solve all PDEs.
Are most differential equations solvable? ›Differential equations such as those used to solve real-life problems may not necessarily be directly solvable, i.e. do not have closed form solutions. Instead, solutions can be approximated using numerical methods. Many fundamental laws of physics and chemistry can be formulated as differential equations.