# First Order Linear Differential Equation

• Prof. 27
In summary, Homework Statement dv/dt = 9.8 - 0.196v is set in correct form. Pauls' Online Notes gets: D*v' = 9.8D. My question is, where did the 0.196 go? I don't see how he's justified in making p(t) just disappear.
Prof. 27

## Homework Statement

dv/dt = 9.8 - 0.196v
Set in correct form:
dv/dt + 0.196v = 9.8
Since p(t) = 0.196, u(t) the integration factor is given by:
u(t) = e∫0.196 dt
Multiply each term by u(t) and rearrange:
(e∫0.196 dt)(dv/dt) + (0.196)(e∫0.196 dt)(v) = (9.8)(e∫0.196 dt)
From now on we will set e∫0.196 dt equal to D and leave it at that.
Here's where I no longer understand it. We take into consideration the product rule:
(f(x)*f(y))' = f'(x)f'(y) + f(x)f'(y)
Pauls' Online Notes gets:
(D*v)' = 9.8D

My question is, where did the 0.196 go? I don't see how he's justified in making p(t) just disappear.

## Homework Equations

(f(x)*f(y))' = f'(x)f'(y) + f(x)f'(y)

## The Attempt at a Solution

My Father who was a math major 30 years ago but went into business and has forgotten a lot
http://tutorial.math.lamar.edu/Classes/DE/Linear.aspx

Last edited:
Prof. 27 said:

## Homework Statement

dv/dt = 9.8 - 0.196v
Set in correct form:
dv/dt + 0.196v = 9.8
Since p(t) = 0.196, u(t) the integration factor is given by:
u(t) = e∫0.196 dt
The right side should be simplified by carrying out the integration. What do you get?
Prof. 27 said:
Multiply each term by u(t) and rearrange:
(e∫0.196 dt)(dv/dt) + (0.196)(e∫0.196 dt)(v) = (9.8)(e∫0.196 dt)
From now on we will set e∫0.196 dt equal to D and leave it at that.
Here's where I no longer understand it. We take into consideration the product rule:
(f(x)*f(y))' = f'(x)f'(y) + f(x)f'(y)
The part above makes no sense. f(x) and f(y) represent the same function, but with different variables. The prime (') represents differentiation, but with respect to which variable?
Prof. 27 said:
Pauls' Online Notes gets:
(D*v)' = 9.8D
Now this makes a little more sense. The idea with an integrating factor is to find a factor that can multiply both sides so that the left side looks like the derivative of a product. What you integrate, you then get just the product.
Prof. 27 said:
My question is, where did the 0.196 go? I don't see how he's justified in making p(t) just disappear.

## Homework Equations

(f(x)*f(y))' = f'(x)f'(y) + f(x)f'(y)

## The Attempt at a Solution

My Father who was a math major 30 years ago but went into business and has forgotten a lot
http://tutorial.math.lamar.edu/Classes/DE/Linear.aspx
A simpler way to do this problem is to use separation. Rewrite the equation as
##\frac{dv}{9.8 - .196v} = dt##
Now integrate.

p(t) gets adsorbed into the integration factor u(t) when the product rule is invoked in step 3 of his solution method.

"The part above makes no sense. f(x) and f(y) represent the same function, but with different variables. The prime (') represents differentiation, but with respect to which variable?"
Sorry, you're right. I should have put g(x) instead of f(y)

Thanks for the help guys, I think I get it now.

Prof. 27 said:

## Homework Statement

dv/dt = 9.8 - 0.196v
Set in correct form:
dv/dt + 0.196v = 9.8
Since p(t) = 0.196, u(t) the integration factor is given by:
u(t) = e∫0.196 dt
Multiply each term by u(t) and rearrange:
(e∫0.196 dt)(dv/dt) + (0.196)(e∫0.196 dt)(v) = (9.8)(e∫0.196 dt)
From now on we will set e∫0.196 dt equal to D and leave it at that.
Here's where I no longer understand it. We take into consideration the product rule:
(f(x)*f(y))' = f'(x)f'(y) + f(x)f'(y)
Pauls' Online Notes gets:
(D*v)' = 9.8D

My question is, where did the 0.196 go? I don't see how he's justified in making p(t) just disappear.

## Homework Equations

(f(x)*f(y))' = f'(x)f'(y) + f(x)f'(y)

## The Attempt at a Solution

My Father who was a math major 30 years ago but went into business and has forgotten a lot
http://tutorial.math.lamar.edu/Classes/DE/Linear.aspx

It is much easier to keep track of things and to trace errors if you write the problem and solution symbolically first, then plug in numbers at the very end.

So, your DE has the form ##dv/dt = a - b v##, and it is just as easy to deal with this as with the version that uses explicit values ##a = 9.8## and ##b = 0.196## at the outset. Note that the RHS has the form ##-b(v - a/b)##, so you will get a simpler problem if you look at ##w = v - a/b## instead of ##v##.

Last edited:

## 1. What is a first order linear differential equation?

A first order linear differential equation is a type of differential equation that involves a function and its first derivative. It can be written in the form y' + P(x)y = Q(x), where P(x) and Q(x) are functions of x. These types of equations are commonly used in mathematical modeling to describe various physical phenomena.

## 2. How do you solve a first order linear differential equation?

To solve a first order linear differential equation, you can use the method of integrating factors. This involves multiplying both sides of the equation by an integrating factor, which is a function that helps to simplify the equation. After multiplying, you can then integrate both sides and solve for the unknown function.

## 3. What is the role of initial conditions in solving a first order linear differential equation?

Initial conditions are necessary when solving a first order linear differential equation because they help to determine the particular solution to the equation. These conditions specify the value of the unknown function at a particular point, and are usually given in the form of y(x0) = y0, where x0 is the starting point and y0 is the corresponding value of the function.

## 4. Can a first order linear differential equation have more than one solution?

Yes, a first order linear differential equation can have more than one solution. This is because the general solution to the equation contains an arbitrary constant, which can take on different values and result in different solutions. However, if initial conditions are given, then the particular solution will be unique.

## 5. How are first order linear differential equations used in real-world applications?

First order linear differential equations are used in various real-world applications, including population growth, radioactive decay, and chemical reactions. They are also used in engineering and physics to model systems and predict future behavior. These equations are essential in understanding and predicting the changes in various physical systems over time.

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