SEPARABLE DIFFERENTIAL EQUATIONS

Objective 4D Solve first and second order separable differential equations, including applications involving initial conditions.

In Unit 2, we explored implicit differentiation, where you find the derivative of a function 𝑦 = 𝑓 𝑥 written as a function of the original function !"

!"= 𝑔(𝑥, 𝑦). This section explores how to find the

original function y given its derivative !"!"

.

A differential equation is an equation that contains an unknown function and one or more of its derivatives. The order of the differential equation is the highest derivative in the equation.

• !"!"= !

! is a ___________________________ order differential equation

• !!!!!!

= −3𝑦! − !! is a ___________________________ order differential equation

• 𝑦! = 𝑥! + 1 is a ___________________________ order differential equation

The solution to a differential equation is a function that satisfies the differential equation when the function and its derivatives are substituted into the equation.

𝑦! = 𝑥!

𝑦 = because

𝑦! = 𝑥!

! = 𝑥!

= 𝑥!

A separable differential equation is a first order differential equation in which the expression for !"!"

can be factored as a function of x times a function of y.

• Which of the following is NOT an example of a separable equation?

𝑑𝑦𝑑𝑥

= 3𝑥!𝑦 𝑦! = 𝑦 3𝑥! + 1 𝑑𝑦𝑑𝑥

= 𝑥! + 2𝑦

Separable equations are nice because they can make antidifferentiation of several types of problems easier. (Note: it will be clearer if you use Leibniz notation – !"

!"– instead of “prime”

notation.)

• Find a general solution of the differential equation !"!"= 2𝑥. (Solve for y)

o Using general integration rules:

o Approaching as a separable differentiable equation:

𝑑𝑦𝑑𝑥

= 2𝑥 Given

“Separate”

Integrate both sides

Simplify

Each integration constant is different – label the left as 𝐶! and the right as 𝐶!

Isolate y

Combine the constants of integration (since it is any arbitrary real number.

• Find a general solution of the differential equation !"

!"= 𝑦.

o This is a function that is its own derivative. What function might that be?

1. Separate

2. Put the y’s on one side and the x’s on the other

3. Rewrite, for integration

4. Integrate both sides

5. Simplify

6. Isolate the y

• Solve !"!"= !!!

!!! for y.

1. Separate

2. Integrate

3. Solve for y

To check any of these, find the derivative(s) of the solution.

Practice. Solve each of the following separable differentiable equations for y.

𝑑𝑦𝑑𝑥

= 𝑦! 2𝑥 − 3 𝑑𝑦𝑑𝑥

= 2𝑥𝑦 + 5𝑦

In applications of differential equations (which we will go over in the next objective), we care less about the family of solutions (general solutions) and more about finding the specific solution that will satisfy an additional requirement. In these cases, we will need to know something about a specific value on the original problem. For example, let’s say we want to know the solution for !"

!!= 2𝑥𝑦 + 5𝑦 and the point 0, 7 .

• Take the general solution you found above.

• Input your given point 0, 7 into the y = ______ equation (the given point here relates to the integral, not the derivative)

• Solve for the constant.

• Replace the constant into your general solution to find the solution that fits this specific criteria.

We most often see these kinds of applications when we are looking for the particular solution that satisfies the initial condition of the form 𝑦 𝑡! = 𝑦!. (For example, given a velocity function, we want to know a particle’s position function, given that it was at y ft at time t.)

Practice. Solve !"

!"= 𝑦! 𝑥! + 6 when 𝑥 = 1 and 𝑦 = 6.

Homework. Find the general solution of each differential equation.

For each problem, find the particular solution of the differential equation that satisfies the initial condition.

1. !"!"= !!

!!

2. !"!"= !

!"#! !

3. !"!"= 3𝑒!!!

4. !"!"= !!

!!!

5. !"!"= !!

!!, 𝑦 2 = 13!

6. !"!"= 2𝑒!!! , 𝑦 −3 = ln !!

!!!!!

7. !"!"= !

!"#! !, 𝑦 3 = 0

8. !"!"= !!

!!, 𝑦 −1 = !!!!!!!

!

9. !"!"= − !

!"#!, 𝑦 3 = !

!

10. !"!"= !!

!!!, 𝑦 2 = !" !

!

11. !"!"= !!!

!!!, 𝑦 1 = 0

12. !"!"= !!!!

!!, 𝑦 −1 = 4!