Differential Equations Basic Concepts Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

What is a Differential Equation? Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

What is a Differential Equation? Short Answer: An equation involving derivatives. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

What is a Differential Equation? Short Answer: An equation involving derivatives. Slightly Longer Answer: When you encounter a dynamic system, where things are changing (especially as time progresses), a differential equation can help model the behavior. A couple of examples: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

What is a Differential Equation? Short Answer: An equation involving derivatives. Slightly Longer Answer: When you encounter a dynamic system, where things are changing (especially as time progresses), a differential equation can help model the behavior. A couple of examples: Here p(t) is a population. It will turn out that the solution is an exponential function, and k is a constant related to the growth rate. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

What is a Differential Equation? Short Answer: An equation involving derivatives. Slightly Longer Answer: When you encounter a dynamic system, where things are changing (especially as time progresses), a differential equation can help model the behavior. A couple of examples: Here p(t) is a population. It will turn out that the solution is an exponential function, and k is a constant related to the growth rate. In this equation y(t) may represent the position of an object that hangs from a spring, driven by an oscillating force. Notice that this equation has a 2nd derivative in it. That makes it a 2nd-order ODE. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Here are a few terms we use to categorize differential equations: Ordinary (ODE) vs. Partial (PDE) Are there any partial derivatives in the DE? YES – it is Partial (a PDE) NO – it is Ordinary (an ODE) We will not solve PDEs in this course. They are generally much harder to solve than ODEs. You may see them later in Math 6B. Order The Order of a DE is the highest derivative that appears. Linear vs. Nonlinear An DE is linear when there are no nonlinear terms of the dependent variable in the equation. i.e. the equation has the form: 𝑎 0 𝑡 𝑦 𝑛 + 𝑎 1 𝑡 𝑦 𝑛−1 +⋯ 𝑎 𝑛−1 𝑡 𝑦 ′ + 𝑎 𝑛 (𝑡)𝑦=𝑔(𝑡) Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

1) 𝑦 ′′ −2𝑥 𝑦 ′ +2𝑝𝑦=0, 𝑝 𝑖𝑠 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 Classify the following DEs by order, linearity, and whether they are ordinary or partial DEs 1) 𝑦 ′′ −2𝑥 𝑦 ′ +2𝑝𝑦=0, 𝑝 𝑖𝑠 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

1) 𝑦 ′′ −2𝑥 𝑦 ′ +2𝑝𝑦=0, 𝑝 𝑖𝑠 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 2nd order, linear ODE Classify the following DEs by order, linearity, and whether they are ordinary or partial DEs 1) 𝑦 ′′ −2𝑥 𝑦 ′ +2𝑝𝑦=0, 𝑝 𝑖𝑠 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 2nd order, linear ODE Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

1) 𝑦 ′′ −2𝑥 𝑦 ′ +2𝑝𝑦=0, 𝑝 𝑖𝑠 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 2nd order, linear ODE Classify the following DEs by order, linearity, and whether they are ordinary or partial DEs 1) 𝑦 ′′ −2𝑥 𝑦 ′ +2𝑝𝑦=0, 𝑝 𝑖𝑠 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 2nd order, linear ODE 2) 𝑦 ′ = 𝑦(2−3𝑥) 𝑥(1+3𝑦) Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

1) 𝑦 ′′ −2𝑥 𝑦 ′ +2𝑝𝑦=0, 𝑝 𝑖𝑠 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 2nd order, linear ODE Classify the following DEs by order, linearity, and whether they are ordinary or partial DEs 1) 𝑦 ′′ −2𝑥 𝑦 ′ +2𝑝𝑦=0, 𝑝 𝑖𝑠 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 2nd order, linear ODE 2) 𝑦 ′ = 𝑦(2−3𝑥) 𝑥(1+3𝑦) 1st order, nonlinear ODE Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

1) 𝑦 ′′ −2𝑥 𝑦 ′ +2𝑝𝑦=0, 𝑝 𝑖𝑠 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 2nd order, linear ODE Classify the following DEs by order, linearity, and whether they are ordinary or partial DEs 1) 𝑦 ′′ −2𝑥 𝑦 ′ +2𝑝𝑦=0, 𝑝 𝑖𝑠 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 2nd order, linear ODE 2) 𝑦 ′ = 𝑦(2−3𝑥) 𝑥(1+3𝑦) 1st order, nonlinear ODE 3) 𝜕 2 𝑢 𝜕 𝑥 2 + 𝜕 2 𝑢 𝜕 𝑦 2 =0 Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

1) 𝑦 ′′ −2𝑥 𝑦 ′ +2𝑝𝑦=0, 𝑝 𝑖𝑠 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 2nd order, linear ODE Classify the following DEs by order, linearity, and whether they are ordinary or partial DEs 1) 𝑦 ′′ −2𝑥 𝑦 ′ +2𝑝𝑦=0, 𝑝 𝑖𝑠 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 2nd order, linear ODE 2) 𝑦 ′ = 𝑦(2−3𝑥) 𝑥(1+3𝑦) 1st order, nonlinear ODE 3) 𝜕 2 𝑢 𝜕 𝑥 2 + 𝜕 2 𝑢 𝜕 𝑦 2 =0 2nd order, linear PDE Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

1) 𝑦 ′′ −2𝑥 𝑦 ′ +2𝑝𝑦=0, 𝑝 𝑖𝑠 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 2nd order, linear ODE Classify the following DEs by order, linearity, and whether they are ordinary or partial DEs 1) 𝑦 ′′ −2𝑥 𝑦 ′ +2𝑝𝑦=0, 𝑝 𝑖𝑠 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 2nd order, linear ODE 2) 𝑦 ′ = 𝑦(2−3𝑥) 𝑥(1+3𝑦) 1st order, nonlinear ODE 3) 𝜕 2 𝑢 𝜕 𝑥 2 + 𝜕 2 𝑢 𝜕 𝑦 2 =0 2nd order, linear PDE 4) 𝑦 1+ 𝑑𝑦 𝑑𝑥 2 =𝐶, 𝐶 𝑖𝑠 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

1) 𝑦 ′′ −2𝑥 𝑦 ′ +2𝑝𝑦=0, 𝑝 𝑖𝑠 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 2nd order, linear ODE Classify the following DEs by order, linearity, and whether they are ordinary or partial DEs 1) 𝑦 ′′ −2𝑥 𝑦 ′ +2𝑝𝑦=0, 𝑝 𝑖𝑠 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 2nd order, linear ODE 2) 𝑦 ′ = 𝑦(2−3𝑥) 𝑥(1+3𝑦) 1st order, nonlinear ODE 3) 𝜕 2 𝑢 𝜕 𝑥 2 + 𝜕 2 𝑢 𝜕 𝑦 2 =0 2nd order, linear PDE 4) 𝑦 1+ 𝑑𝑦 𝑑𝑥 2 =𝐶, 𝐶 𝑖𝑠 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 1st order, nonlinear ODE Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

1) 𝑦 ′′ −2𝑥 𝑦 ′ +2𝑝𝑦=0, 𝑝 𝑖𝑠 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 2nd order, linear ODE Classify the following DEs by order, linearity, and whether they are ordinary or partial DEs 1) 𝑦 ′′ −2𝑥 𝑦 ′ +2𝑝𝑦=0, 𝑝 𝑖𝑠 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 2nd order, linear ODE 2) 𝑦 ′ = 𝑦(2−3𝑥) 𝑥(1+3𝑦) 1st order, nonlinear ODE 3) 𝜕 2 𝑢 𝜕 𝑥 2 + 𝜕 2 𝑢 𝜕 𝑦 2 =0 2nd order, linear PDE 4) 𝑦 1+ 𝑑𝑦 𝑑𝑥 2 =𝐶, 𝐶 𝑖𝑠 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 1st order, nonlinear ODE 5) 5 𝑥 ′′ +2 𝑥 ′ +9𝑥=2cos⁡(3𝑡) Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

1) 𝑦 ′′ −2𝑥 𝑦 ′ +2𝑝𝑦=0, 𝑝 𝑖𝑠 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 2nd order, linear ODE Classify the following DEs by order, linearity, and whether they are ordinary or partial DEs 1) 𝑦 ′′ −2𝑥 𝑦 ′ +2𝑝𝑦=0, 𝑝 𝑖𝑠 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 2nd order, linear ODE 2) 𝑦 ′ = 𝑦(2−3𝑥) 𝑥(1+3𝑦) 1st order, nonlinear ODE 3) 𝜕 2 𝑢 𝜕 𝑥 2 + 𝜕 2 𝑢 𝜕 𝑦 2 =0 2nd order, linear PDE 4) 𝑦 1+ 𝑑𝑦 𝑑𝑥 2 =𝐶, 𝐶 𝑖𝑠 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 1st order, nonlinear ODE 5) 5 𝑥 ′′ +2 𝑥 ′ +9𝑥=2cos⁡(3𝑡) 2nd order, linear ODE Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Let’s take a look at the population growth model. Suppose that the population of a country is initially 2 million people. If we make the simple assumption that the population will grow at a rate of 1% per year we arrive at this equation: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Let’s take a look at the population growth model. Suppose that the population of a country is initially 2 million people. If we make the simple assumption that the population will grow at a rate of 1% per year we arrive at this equation: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Let’s take a look at the population growth model. Suppose that the population of a country is initially 2 million people. If we make the simple assumption that the population will grow at a rate of 1% per year we arrive at this equation: We will come across several ways to solve this equation. An explicit solution (a function p(t)) can be found via an educated guess, an integrating factor or a method we will call separation of variables. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Let’s take a look at the population growth model. Suppose that the population of a country is initially 2 million people. If we make the simple assumption that the population will grow at a rate of 1% per year we arrive at this equation: We will come across several ways to solve this equation. An explicit solution (a function p(t)) can be found via an educated guess, an integrating factor or a method we will call separation of variables. An approximate solution, which would give us an approximate population figure for any particular year, can be found via Euler’s method (or some more sophisticated numerical method). Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Suppose that the population of a country is initially 2 million people Suppose that the population of a country is initially 2 million people. If we make the simple assumption that the population will grow at a rate of 1% per year we arrive at this equation: First let’s draw a SLOPE FIELD for this ODE. This is a graph where at every point we can put an arrow indicating the SLOPE of the solution to the ODE. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

For example, at the point (t=0, p=0) we get dp/dt = 0. Suppose that the population of a country is initially 2 million people. If we make the simple assumption that the population will grow at a rate of 1% per year we arrive at this equation: First let’s draw a SLOPE FIELD for this ODE. This is a graph where at every point we can put an arrow indicating the SLOPE of the solution to the ODE. We can accomplish this by simply plugging values into the equation and calculating values of the derivative dp/dt. For example, at the point (t=0, p=0) we get dp/dt = 0. At the point (t=0, p=100) we get dp/dt = 1. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

For example, at the point (t=0, p=0) we get dp/dt = 0. Suppose that the population of a country is initially 2 million people. If we make the simple assumption that the population will grow at a rate of 1% per year we arrive at this equation: First let’s draw a SLOPE FIELD for this ODE. This is a graph where at every point we can put an arrow indicating the SLOPE of the solution to the ODE. We can accomplish this by simply plugging values into the equation and calculating values of the derivative dp/dt. For example, at the point (t=0, p=0) we get dp/dt = 0. At the point (t=0, p=100) we get dp/dt = 1. We can do this all day and get a nice picture, or we can be a bit more observant and notice that there is no ‘t’ in our equation. This means that our slopes will be independent of t; they only depend on the value of p. This shortens our work considerably. Of course we could also ask our computer to graph it for us… Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Here is the slope field, along with a few solution curves. Suppose that the population of a country is initially 2 million people. If we make the simple assumption that the population will grow at a rate of 1% per year we arrive at this equation: Here is the slope field, along with a few solution curves. This curve solves our IVP. This curve is the equilibrium solution p(t)=0. This curve goes through p(0)=-1. Mathematically it is a valid solution, but it does not make physical sense (the population is never negative!) Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

We can arrive at an explicit formula solution by separating variables: Suppose that the population of a country is initially 2 million people. If we make the simple assumption that the population will grow at a rate of 1% per year we arrive at this equation: We can arrive at an explicit formula solution by separating variables: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

We can arrive at an explicit formula solution by separating variables: Suppose that the population of a country is initially 2 million people. If we make the simple assumption that the population will grow at a rate of 1% per year we arrive at this equation: We can arrive at an explicit formula solution by separating variables: Here is our explicit solution. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Let’s verify that this function solves the DE. Suppose that the population of a country is initially 2 million people. If we make the simple assumption that the population will grow at a rate of 1% per year we arrive at this equation: Let’s verify that this function solves the DE. We need to find the derivative, and then see that it fits the equation. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Let’s verify that this function solves the DE. Suppose that the population of a country is initially 2 million people. If we make the simple assumption that the population will grow at a rate of 1% per year we arrive at this equation: Let’s verify that this function solves the DE. We need to find the derivative, and then see that it fits the equation. Yep, this is exactly 0.01p Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Let’s verify that this function solves the DE. Suppose that the population of a country is initially 2 million people. If we make the simple assumption that the population will grow at a rate of 1% per year we arrive at this equation: Let’s verify that this function solves the DE. We need to find the derivative, and then see that it fits the equation. Yep, this is exactly 0.01p We also need to know that the function matches the given initial value. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Let’s verify that this function solves the DE. Suppose that the population of a country is initially 2 million people. If we make the simple assumption that the population will grow at a rate of 1% per year we arrive at this equation: Let’s verify that this function solves the DE. We need to find the derivative, and then see that it fits the equation. Yep, this is exactly 0.01p We also need to know that the function matches the given initial value. Yep, this matches up. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Verify that the given function satisfies the differential equation.     Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Verify that the given function satisfies the differential equation.     First compute the derivative:   Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

 This is true, so the diff. eq. is satisfied Verify that the given function satisfies the differential equation.     First compute the derivative:     Now substitute into the differential equation:    This is true, so the diff. eq. is satisfied (there is no initial condition to satisfy, so we are done) Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Verify that the given function satisfies the differential equation, and find a value for C that satisfies the given initial condition. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Verify that the given function satisfies the differential equation, and find a value for C that satisfies the given initial condition. Find the derivative:   Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Verify that the given function satisfies the differential equation, and find a value for C that satisfies the given initial condition. Find the derivative:   Substitute into diff. eq.: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

 It all cancels out, so the equation is satisfied Verify that the given function satisfies the differential equation, and find a value for C that satisfies the given initial condition. Find the derivative:   Substitute into diff. eq.:  It all cancels out, so the equation is satisfied Now we need to satisfy the initial condition. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

 It all cancels out, so the equation is satisfied Verify that the given function satisfies the differential equation, and find a value for C that satisfies the given initial condition. Find the derivative:   Substitute into diff. eq.:  It all cancels out, so the equation is satisfied Now we need to satisfy the initial condition. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

 It all cancels out, so the equation is satisfied Verify that the given function satisfies the differential equation, and find a value for C that satisfies the given initial condition. Find the derivative:   Substitute into diff. eq.:  It all cancels out, so the equation is satisfied Now we need to satisfy the initial condition. Set this equal to -1 Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

 It all cancels out, so the equation is satisfied Verify that the given function satisfies the differential equation, and find a value for C that satisfies the given initial condition. Find the derivative:   Substitute into diff. eq.:  It all cancels out, so the equation is satisfied Now we need to satisfy the initial condition. Set this equal to -1 Our final solution is Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB