ODE - First encounter, not understanding them

[Astrum]

Homework Statement

First time I've had to deal with ODEs, an I'm pretty confused.

This SHOULD be a simple ODE for finding air resistance, that is only dealing with the y vector (up and down in this case)

[itex]m\frac{dv_{y}}{dt}=mg-kv_{y}[/itex]

Homework Equations

F=ma
f=-kV

The Attempt at a Solution

First order differential equation, so we only need to integrate once.

[tex]m\frac{dv_{y}}{dt}=mg-kv_{y}=g-\frac{kv}{m}[/tex]

[tex]\int\frac{dv}{dt}=gt-\frac{k}{m}\int v dt[/tex]

I have no idea where to go from here. The answer in the book gives:

[tex]v=\frac{mg}{k}(1-e^{-\frac{kt}{m}})[/tex]

 

[Mark44]

Homework Statement

First time I've had to deal with ODEs, an I'm pretty confused.

This SHOULD be a simple ODE for finding air resistance, that is only dealing with the y vector (up and down in this case)

[itex]m\frac{dv_{y}}{dt}=mg-kv_{y}[/itex]

Homework Equations

F=ma
f=-kV

The Attempt at a Solution

First order differential equation, so we only need to integrate once.

[tex]m\frac{dv_{y}}{dt}=mg-kv_{y}=g-\frac{kv}{m}[/tex]

[tex]\int\frac{dv}{dt}=gt-\frac{k}{m}\int v dt[/tex]

This (above) won't work.

I have no idea where to go from here. The answer in the book gives:

[tex]v=\frac{mg}{k}(1-e^{-\frac{kt}{m}})[/tex]

Divide both sides by m to get
dv/dt = g - (k/m)v

This equation is separable, so get everything involving v and dv on one side, and everything involving t and dt on the other. (There is nothing that involves t, though.)

 

[rude man]

The book answer does not include a possible initial velocity .

I'm sure there's a "classical" way to solve this equation. As an EE I would of course use the Laplace transform method, which sadly is not taught in most diff. eq. courses but should be.

The book answer is correct if initial velocity = 0.

(If you're interested in the Laplace method let me know.)

EDIT: Mark44 has the way to go. I tried but could not figure out how to separate the variables.

 

[Astrum]

This (above) won't work.Divide both sides by m to get
dv/dt = g - (k/m)v

This equation is separable, so get everything involving v and dv on one side, and everything involving t and dt on the other. (There is nothing that involves t, though.)

I'll be honest, I had to look up what a separable ODE was.

So, after doing that, I got:

[tex]\int\frac{k}{m}v-g dv=\int dt[/tex]

so this equals: [itex]\frac{k}{2m}v^{2}-gv=t[/itex]

And the next step is to just rearrange?

 

[Mark44]

I'll be honest, I had to look up what a separable ODE was.

So, after doing that, I got:

[tex]\int\frac{k}{m}v-g dv=\int dt[/tex]

This isn't right, due to an algebra error.

Starting from dv/dt = g - (k/m)v,

we have dv/(g - (k/m)v) = dt

Now the equation is separated, and you can integrate both sides.

so this equals: [itex]\frac{k}{2m}v^{2}-gv=t[/itex]

And the next step is to just rearrange?

 

[Mark44]

The book answer does not include a possible initial velocity .

I'm sure there's a "classical" way to solve this equation. As an EE I would of course use the Laplace transform method, which sadly is not taught in most diff. eq. courses but should be.

I taught diff. eqns. for a lot of years, and Laplace transforms were one of the techniques presented. They weren't taught at the beginning of the course, though, and that's where the OP is, I believe.

The book answer is correct if initial velocity = 0.

(If you're interested in the Laplace method let me know.)

EDIT: Mark44 has the way to go. I tried but could not figure out how to separate the variables.

 

[rude man]

I taught diff. eqns. for a lot of years, and Laplace transforms were one of the techniques presented. They weren't taught at the beginning of the course, though, and that's where the OP is, I believe.

I had two texts on diff. eq's and neither mentioned Laplace. One was by MIT people.

 

[HallsofIvy]

Personally, I dislike the "Laplace transform method". It loved by engineers because it gives a very "mechanical" way of solving differential equations. You just "look up" the inverse transfors in a table. But I have never met a differentia equation, solvable by Laplace transform, that was not, in my opinion, more easily solved by direct methods.

 

[rude man]

Personally, I dislike the "Laplace transform method". It loved by engineers because it gives a very "mechanical" way of solving differential equations. You just "look up" the inverse transfors in a table. But I have never met a differentia equation, solvable by Laplace transform, that was not, in my opinion, more easily solved by direct methods.

I'm not sure that "guessing" a solution to a diff. eq., as is done classically, is more insightful than looking up an inverse in a table. And I like the convenience of not having to do two solutions (homogeneous & inhomogeneous) and also including the initial conditions automatically.

Chaqu'un a son gout.