1. Differential Equations
Basic Concepts
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2. What is a Differential Equation?
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3. What is a Differential Equation?
Short Answer: An equation involving derivatives.
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4. 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:
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5. 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.
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6. 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.
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7. 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)
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8. 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
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9. 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
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10. 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)
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11. 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
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12. 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 𝑢 𝜕 𝑦 2 =0
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13. 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 𝑢 𝜕 𝑦 2 =0
2nd order, linear PDE
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14. 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 𝑢 𝜕 𝑦 2 =0
2nd order, linear PDE
4) 𝑦 1+ 𝑑𝑦 𝑑𝑥 2 =𝐶, 𝐶 𝑖𝑠 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
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15. 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 𝑢 𝜕 𝑦 2 =0
2nd order, linear PDE
4) 𝑦 1+ 𝑑𝑦 𝑑𝑥 2 =𝐶, 𝐶 𝑖𝑠 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
1st order, nonlinear ODE
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16. 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 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)
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17. 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 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
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18. 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:
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19. 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:
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20. 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.
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21. 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).
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22. 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.
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23. 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.
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24. 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…
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25. 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!)
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26. 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:
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27. 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.
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28. 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.
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29. 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
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30. 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.
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31. 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.
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32. Verify that the given function satisfies the differential equation.
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33. Verify that the given function satisfies the differential equation.
First compute the derivative:
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34.  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)
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35. Verify that the given function satisfies the differential equation,
and find a value for C that satisfies the given initial condition.
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36. Verify that the given function satisfies the differential equation,
and find a value for C that satisfies the given initial condition.
Find the derivative:
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37. 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.:
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38.  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.
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39.  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.
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40.  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
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41.  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
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