9.1/9.3 – Differential Equations Math 181 9.1/9.3 – Differential Equations

A differential equation is an equation that contains an unknown function and some of its derivatives. The following are examples of differential equations: 𝑑 2 𝑦 𝑑 𝑡 2 =𝑔 𝑚 𝑣 ′ =𝑚𝑔−𝑏 𝑣 2 𝑚 𝑦 ′′ +𝑘𝑦=0 𝐿 𝜃 ′′ +𝑔 sin 𝜃 =0

A differential equation is an equation that contains an unknown function and some of its derivatives. The following are examples of differential equations: 𝑑 2 𝑦 𝑑 𝑡 2 =𝑔 𝑚 𝑣 ′ =𝑚𝑔−𝑏 𝑣 2 𝐿 𝜃 ′′ +𝑔 sin 𝜃 =0

Population growth: 𝑑𝑃 𝑑𝑡 =𝑘𝑃 More examples: Population growth: 𝑑𝑃 𝑑𝑡 =𝑘𝑃 (the more people there are, the higher the birthrate) Motion of spring: 𝑚 𝑑 2 𝑥 𝑑 𝑡 2 =−𝑘𝑥 (recall Hooke’s law was 𝐹=𝑘𝑥)

Differential equations are abbreviated “DEs Differential equations are abbreviated “DEs.” A solution to a DE is a function. The order of a DE is the highest derivative that appears in the equation.

Ex 1. Show that 𝑦=𝑥 sin 2𝑥 is a solution to the following differential equation. 𝑦 ′ − 𝑦 𝑥 =2𝑥 cos 2𝑥

Ex 2. Show that each member of the family of functions 𝑦=𝐶 𝑒 3𝑥 is a solution to the following differential equation. 𝑦 ′ =3𝑦

The general solution to a DE is a family of solutions, as in example 2 above. But if we’re given an initial condition (for example, 𝑦 0 =4), then the particular solution of both the DE and initial condition is the function that satisfies both the DE and the initial condition. A combination of a DE and an initial condition is called an initial-value problem.

The general solution to a DE is a family of solutions, as in example 2 above. But if we’re given an initial condition (for example, 𝑦 0 =4), then the particular solution of both the DE and initial condition is the function that satisfies both the DE and the initial condition. A combination of a DE and an initial condition is called an initial-value problem.

The general solution to a DE is a family of solutions, as in example 2 above. But if we’re given an initial condition (for example, 𝑦 0 =4), then the particular solution of both the DE and initial condition is the function that satisfies both the DE and the initial condition. A combination of a DE and an initial condition is called an initial-value problem.

For example, “ 𝑦 ′ =3𝑦, 𝑦 0 =4” is an initial-value problem For example, “ 𝑦 ′ =3𝑦, 𝑦 0 =4” is an initial-value problem. 𝑦=4 𝑒 3𝑥 is a particular solution to this initial-value problem because it is a solution of both 𝑦 ′ =3𝑦 and 𝑦 0 =4.

For example, “ 𝑦 ′ =3𝑦, 𝑦 0 =4” is an initial-value problem For example, “ 𝑦 ′ =3𝑦, 𝑦 0 =4” is an initial-value problem. 𝑦=4 𝑒 3𝑥 is a particular solution to this initial-value problem because it is a solution of both 𝑦 ′ =3𝑦 and 𝑦 0 =4.

What will the solutions of 𝑦 ′ =𝑓(𝑥,𝑦) look like?

Notice that for each 𝑥,𝑦 point on the plane, 𝑑𝑦 𝑑𝑥 =𝑓 𝑥,𝑦 gives us the slope of the tangent line of a solution curve through that point. We can fill the plane with these little tangent “stubs” to generate what is called a __________ (or a _____________).

Notice that for each 𝑥,𝑦 point on the plane, 𝑑𝑦 𝑑𝑥 =𝑓 𝑥,𝑦 gives us the slope of the tangent line of a solution curve through that point. We can fill the plane with these little tangent “stubs” to generate what is called a slope field (or a direction field).

𝒚 ′ =𝒚− 𝒙 𝟐

𝒚 ′ =𝒚+ 𝐬𝐢𝐧 𝒙

Solutions curves “flow” with the tangent stubs. 𝒚 ′ =𝒚+ 𝐬𝐢𝐧 𝒙

Ex 3. Without making a slope field, what can you say about any given solution to 𝑦 ′ =− 𝑦 4 ?

Slope field for 𝒚 ′ =− 𝒚 𝟒

A differential equation 𝑑𝑦 𝑑𝑥 =𝑓(𝑥,𝑦) is separable if you can write it in the form: 𝑑𝑦 𝑑𝑥 = 𝑔 𝑥 ℎ 𝑦 In this case, we can take the following steps to solve the DE: ℎ 𝑦 𝑑𝑦=𝑔 𝑥 𝑑𝑥 ℎ 𝑦 𝑑𝑦= 𝑔 𝑥 𝑑𝑥

A differential equation 𝑑𝑦 𝑑𝑥 =𝑓(𝑥,𝑦) is separable if you can write it in the form: 𝑑𝑦 𝑑𝑥 = 𝑔 𝑥 ℎ 𝑦 In this case, we can take the following steps to solve the DE: ℎ 𝑦 𝑑𝑦=𝑔 𝑥 𝑑𝑥 ℎ 𝑦 𝑑𝑦= 𝑔 𝑥 𝑑𝑥

A differential equation 𝑑𝑦 𝑑𝑥 =𝑓(𝑥,𝑦) is separable if you can write it in the form: 𝑑𝑦 𝑑𝑥 = 𝑔 𝑥 ℎ 𝑦 In this case, we can take the following steps to solve the DE: ℎ 𝑦 𝑑𝑦=𝑔 𝑥 𝑑𝑥 ℎ 𝑦 𝑑𝑦= 𝑔 𝑥 𝑑𝑥

A differential equation 𝑑𝑦 𝑑𝑥 =𝑓(𝑥,𝑦) is separable if you can write it in the form: 𝑑𝑦 𝑑𝑥 = 𝑔 𝑥 ℎ 𝑦 In this case, we can take the following steps to solve the DE: ℎ 𝑦 𝑑𝑦=𝑔 𝑥 𝑑𝑥 ℎ 𝑦 𝑑𝑦= 𝑔 𝑥 𝑑𝑥

Ex 4. Find the general solution to the following differential equation

Ex 5. Find the general solution to the following differential equation

An orthogonal trajectory is of a family of curves is a curve that intersects each curve of the family at right angles (that is, orthogonally).

What do you think the orthogonal trajectory of the family of parabolas (see right) going through (1, 0.5) looks like?

Ex 6. Find the orthogonal trajectories of the family of curves 𝑥𝑦=𝑎, where 𝑎≠0 is an arbitrary constant.

Mixing Problems If 𝐴 measures the amount of stuff (liquid, gas, population, etc.) in a compartment, then 𝑑𝐴 𝑑𝑡 =input rate−output rate

Mixing Problems Ex 7. A big tank initially contains 1000 L of a salt-water solution, which has a concentration of 0.01 kg of salt per liter. Brine (a solution of salt in water) containing 0.03 kg of salt per liter of water pours into the tank at 15 L/min. Assume the solution in the tank is kept well-mixed and drains from the tank at 15 L/min. How much salt is in the tank initially? How many liters of solution are in the tank at any given time? Assuming the amount of salt in the tank is 𝐴 𝑡 , what is the concentration of salt in the tank at time 𝑡? What is the rate at which salt is entering into the tank? What is the rate at which salt is leaving the tank? How much salt is in the tank after 10 minutes?

Mixing Problems Ex 7. A big tank initially contains 1000 L of a salt-water solution, which has a concentration of 0.01 kg of salt per liter. Brine (a solution of salt in water) containing 0.03 kg of salt per liter of water pours into the tank at 15 L/min. Assume the solution in the tank is kept well-mixed and drains from the tank at 15 L/min. How much salt is in the tank initially? How many liters of solution are in the tank at any given time? Assuming the amount of salt in the tank is 𝐴 𝑡 , what is the concentration of salt in the tank at time 𝑡? What is the rate at which salt is entering into the tank? What is the rate at which salt is leaving the tank? How much salt is in the tank after 10 minutes?

Mixing Problems Ex 7. A big tank initially contains 1000 L of a salt-water solution, which has a concentration of 0.01 kg of salt per liter. Brine (a solution of salt in water) containing 0.03 kg of salt per liter of water pours into the tank at 15 L/min. Assume the solution in the tank is kept well-mixed and drains from the tank at 15 L/min. How much salt is in the tank initially? How many liters of solution are in the tank at any given time? Assuming the amount of salt in the tank is 𝐴 𝑡 , what is the concentration of salt in the tank at time 𝑡? What is the rate at which salt is entering into the tank? What is the rate at which salt is leaving the tank? How much salt is in the tank after 10 minutes?

Mixing Problems Ex 7. A big tank initially contains 1000 L of a salt-water solution, which has a concentration of 0.01 kg of salt per liter. Brine (a solution of salt in water) containing 0.03 kg of salt per liter of water pours into the tank at 15 L/min. Assume the solution in the tank is kept well-mixed and drains from the tank at 15 L/min. How much salt is in the tank initially? How many liters of solution are in the tank at any given time? Assuming the amount of salt in the tank is 𝐴 𝑡 , what is the concentration of salt in the tank at time 𝑡? What is the rate at which salt is entering into the tank? What is the rate at which salt is leaving the tank? How much salt is in the tank after 10 minutes?

Mixing Problems Ex 7. A big tank initially contains 1000 L of a salt-water solution, which has a concentration of 0.01 kg of salt per liter. Brine (a solution of salt in water) containing 0.03 kg of salt per liter of water pours into the tank at 15 L/min. Assume the solution in the tank is kept well-mixed and drains from the tank at 15 L/min. How much salt is in the tank initially? How many liters of solution are in the tank at any given time? Assuming the amount of salt in the tank is 𝐴 𝑡 , what is the concentration of salt in the tank at time 𝑡? What is the rate at which salt is entering into the tank? What is the rate at which salt is leaving the tank? How much salt is in the tank after 10 minutes?

Mixing Problems Ex 7. A big tank initially contains 1000 L of a salt-water solution, which has a concentration of 0.01 kg of salt per liter. Brine (a solution of salt in water) containing 0.03 kg of salt per liter of water pours into the tank at 15 L/min. Assume the solution in the tank is kept well-mixed and drains from the tank at 15 L/min. How much salt is in the tank initially? How many liters of solution are in the tank at any given time? Assuming the amount of salt in the tank is 𝐴 𝑡 , what is the concentration of salt in the tank at time 𝑡? What is the rate at which salt is entering into the tank? What is the rate at which salt is leaving the tank? How much salt is in the tank after 10 minutes?

We can justify our procedure for solving separable DEs using the Chain Rule. If you start with ℎ 𝑦 𝑑𝑦= 𝑔 𝑥 𝑑𝑥, then 𝑑 𝑑𝑥 ℎ 𝑦 𝑑𝑦 = 𝑑 𝑑𝑥 𝑔 𝑥 𝑑𝑥 𝑑 𝑑𝑦 ℎ 𝑦 𝑑𝑦 ⋅ 𝑑𝑦 𝑑𝑥 = 𝑑 𝑑𝑥 𝑔 𝑥 𝑑𝑥 (𝑏𝑦 𝑡ℎ𝑒 𝐶ℎ𝑎𝑖𝑛 𝑅𝑢𝑙𝑒) ℎ 𝑦 ⋅ 𝑑𝑦 𝑑𝑥 =𝑔 𝑥 (𝑏𝑦 𝑡ℎ𝑒 𝐹𝑢𝑛𝑑𝑎𝑚𝑒𝑛𝑡𝑎𝑙 𝑇ℎ𝑒𝑜𝑟𝑒𝑚 𝑜𝑓 𝐶𝑎𝑙𝑐𝑢𝑙𝑢𝑠) 𝑑𝑦 𝑑𝑥 = 𝑔 𝑥 ℎ 𝑦 (𝑏𝑦 𝐽𝑜𝑣𝑒, 𝑤 𝑒 ′ 𝑣𝑒 𝑑𝑜𝑛𝑒 𝑖𝑡!)

We can justify our procedure for solving separable DEs using the Chain Rule. If you start with ℎ 𝑦 𝑑𝑦= 𝑔 𝑥 𝑑𝑥, then 𝑑 𝑑𝑥 ℎ 𝑦 𝑑𝑦 = 𝑑 𝑑𝑥 𝑔 𝑥 𝑑𝑥 𝑑 𝑑𝑦 ℎ 𝑦 𝑑𝑦 ⋅ 𝑑𝑦 𝑑𝑥 = 𝑑 𝑑𝑥 𝑔 𝑥 𝑑𝑥 (𝑏𝑦 𝑡ℎ𝑒 𝐶ℎ𝑎𝑖𝑛 𝑅𝑢𝑙𝑒) ℎ 𝑦 ⋅ 𝑑𝑦 𝑑𝑥 =𝑔 𝑥 (𝑏𝑦 𝑡ℎ𝑒 𝐹𝑢𝑛𝑑𝑎𝑚𝑒𝑛𝑡𝑎𝑙 𝑇ℎ𝑒𝑜𝑟𝑒𝑚 𝑜𝑓 𝐶𝑎𝑙𝑐𝑢𝑙𝑢𝑠) 𝑑𝑦 𝑑𝑥 = 𝑔 𝑥 ℎ 𝑦 (𝑏𝑦 𝐽𝑜𝑣𝑒, 𝑤 𝑒 ′ 𝑣𝑒 𝑑𝑜𝑛𝑒 𝑖𝑡!)

We can justify our procedure for solving separable DEs using the Chain Rule. If you start with ℎ 𝑦 𝑑𝑦= 𝑔 𝑥 𝑑𝑥, then 𝑑 𝑑𝑥 ℎ 𝑦 𝑑𝑦 = 𝑑 𝑑𝑥 𝑔 𝑥 𝑑𝑥 𝑑 𝑑𝑦 ℎ 𝑦 𝑑𝑦 ⋅ 𝑑𝑦 𝑑𝑥 = 𝑑 𝑑𝑥 𝑔 𝑥 𝑑𝑥 (𝑏𝑦 𝑡ℎ𝑒 𝐶ℎ𝑎𝑖𝑛 𝑅𝑢𝑙𝑒) ℎ 𝑦 ⋅ 𝑑𝑦 𝑑𝑥 =𝑔 𝑥 (𝑏𝑦 𝑡ℎ𝑒 𝐹𝑢𝑛𝑑𝑎𝑚𝑒𝑛𝑡𝑎𝑙 𝑇ℎ𝑒𝑜𝑟𝑒𝑚 𝑜𝑓 𝐶𝑎𝑙𝑐𝑢𝑙𝑢𝑠) 𝑑𝑦 𝑑𝑥 = 𝑔 𝑥 ℎ 𝑦 (𝑏𝑦 𝐽𝑜𝑣𝑒, 𝑤 𝑒 ′ 𝑣𝑒 𝑑𝑜𝑛𝑒 𝑖𝑡!)

We can justify our procedure for solving separable DEs using the Chain Rule. If you start with ℎ 𝑦 𝑑𝑦= 𝑔 𝑥 𝑑𝑥, then 𝑑 𝑑𝑥 ℎ 𝑦 𝑑𝑦 = 𝑑 𝑑𝑥 𝑔 𝑥 𝑑𝑥 𝑑 𝑑𝑦 ℎ 𝑦 𝑑𝑦 ⋅ 𝑑𝑦 𝑑𝑥 = 𝑑 𝑑𝑥 𝑔 𝑥 𝑑𝑥 (𝑏𝑦 𝑡ℎ𝑒 𝐶ℎ𝑎𝑖𝑛 𝑅𝑢𝑙𝑒) ℎ 𝑦 ⋅ 𝑑𝑦 𝑑𝑥 =𝑔 𝑥 (𝑏𝑦 𝑡ℎ𝑒 𝐹𝑢𝑛𝑑𝑎𝑚𝑒𝑛𝑡𝑎𝑙 𝑇ℎ𝑒𝑜𝑟𝑒𝑚 𝑜𝑓 𝐶𝑎𝑙𝑐𝑢𝑙𝑢𝑠) 𝑑𝑦 𝑑𝑥 = 𝑔 𝑥 ℎ 𝑦 (𝑏𝑦 𝐽𝑜𝑣𝑒, 𝑤 𝑒 ′ 𝑣𝑒 𝑑𝑜𝑛𝑒 𝑖𝑡!)

We can justify our procedure for solving separable DEs using the Chain Rule. If you start with ℎ 𝑦 𝑑𝑦= 𝑔 𝑥 𝑑𝑥, then 𝑑 𝑑𝑥 ℎ 𝑦 𝑑𝑦 = 𝑑 𝑑𝑥 𝑔 𝑥 𝑑𝑥 𝑑 𝑑𝑦 ℎ 𝑦 𝑑𝑦 ⋅ 𝑑𝑦 𝑑𝑥 = 𝑑 𝑑𝑥 𝑔 𝑥 𝑑𝑥 (𝑏𝑦 𝑡ℎ𝑒 𝐶ℎ𝑎𝑖𝑛 𝑅𝑢𝑙𝑒) ℎ 𝑦 ⋅ 𝑑𝑦 𝑑𝑥 =𝑔 𝑥 (𝑏𝑦 𝑡ℎ𝑒 𝐹𝑢𝑛𝑑𝑎𝑚𝑒𝑛𝑡𝑎𝑙 𝑇ℎ𝑒𝑜𝑟𝑒𝑚 𝑜𝑓 𝐶𝑎𝑙𝑐𝑢𝑙𝑢𝑠) 𝑑𝑦 𝑑𝑥 = 𝑔 𝑥 ℎ 𝑦 (𝑏𝑦 𝐽𝑜𝑣𝑒, 𝑤 𝑒 ′ 𝑣𝑒 𝑑𝑜𝑛𝑒 𝑖𝑡!)