First Order Linear Differential Equation

[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

 

[Mark44]

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?

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?

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.

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.

 

[paisiello2]

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

 

[Prof. 27]

"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.

 

[Ray Vickson]

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##.