Differential equations for dummies...

[Sukhoi]

Does anyone know of any books or websites that explain diffeqs in very basic terms? 🙂 Something with them just isn't clicking (mostly v-substitution), and I bombed my test today. 🙁

 

[Hitman32]

A differential equation is where there is a variable that represents the rate of change of something. So instead of y=f(x) you have y'=f(x). There are various types of differential equations, some of them being seperable, meaning you can just integrate both sides to put the equation in algebraic terms while others are inseperable etc.

 

[Syringer]

Originally posted by: Sukhoi
Does anyone know of any books or websites that explain diffeqs in very basic terms? 🙂 Something with them just isn't clicking (mostly v-substitution), and I bombed my test today. 🙁

I actually just had a test today..and I hope I did well on it. If you have any questions just post them here and I'm sure someone will answer it. To me it's actually a lot easier than Calculus II and Multivariable Calc..

 

[Sukhoi]

Right, I know the basic stuff. I can do separable equations and linear first-order just fine. I'm having problems with anything more complex than that (we haven't gotten all that far).

 

[Sukhoi]

Originally posted by: Syringer

Originally posted by: Sukhoi
Does anyone know of any books or websites that explain diffeqs in very basic terms? 🙂 Something with them just isn't clicking (mostly v-substitution), and I bombed my test today. 🙁

I actually just had a test today..and I hope I did well on it. If you have any questions just post them here and I'm sure someone will answer it. To me it's actually a lot easier than Calculus II and Multivariable Calc..

Can anyone either explain or point me to a site that explains step by step how to do the v-substitution for first-order equations? It's quite confusing for me because my textbook and my teacher do it in a different order. I thought I had it figured out last night, but of course I couldn't do it on the test today. Though I suppose it didn't help that our test seemed to be quite a bit harder than the practice exam.

 

[RaynorWolfcastle]

what the hell is v-substitution? is it the method for solving exact equations or Bernoulli eqns perhaps? I could help if I knew what kind of eqn you're talking about.

Since I'm going to bed right now I'll point you to my old prof's notes

*Beware* this is a 150 page pdf
linkage

 

[Syringer]

Hmm, I'm not sure about v substitutions, but we've done u subs which is probably the same thing if you're still in the early stages.

Basically you have an equation that's in the form:

M(x,y) dx + N(x,y) dy = 0

All you have to do is do two substitutions:

let y = ux (or x = uy if it's easier to integrate the M function than N)

and let dy = x du + u dx (via product rule)

and just go ahead and plug it all in..simplify as much as possible, and then you should have a separable equation in terms of u and x, or u and y. Integrate it as usual, then plug y/x for all u's back in (or x/y). If you have a sample problem I can help you work through it..

 

[RaynorWolfcastle]

Originally posted by: Syringer
Hmm, I'm not sure about v substitutions, but we've done u subs which is probably the same thing if you're still in the early stages.

Basically you have an equation that's in the form:

M(x,y) dx + N(x,y) dy = 0

All you have to do is do two substitutions:

let y = ux (or x = uy if it's easier to integrate the M function than N)

and let dy = x du + u dx (via product rule)

and just go ahead and plug it all in..simplify as much as possible, and then you should have a separable equation in terms of u and x, or u and y. Integrate it as usual, then plug y/x for all u's back in (or x/y). If you have a sample problem I can help you work through it..

you're describing an exact equation 😉

 

[onza]

Originally posted by: Syringer
Hmm, I'm not sure about v substitutions, but we've done u subs which is probably the same thing if you're still in the early stages.

Basically you have an equation that's in the form:

M(x,y) dx + N(x,y) dy = 0

All you have to do is do two substitutions:

let y = ux (or x = uy if it's easier to integrate the M function than N)

and let dy = x du + u dx (via product rule)

and just go ahead and plug it all in..simplify as much as possible, and then you should have a separable equation in terms of u and x, or u and y. Integrate it as usual, then plug y/x for all u's back in (or x/y). If you have a sample problem I can help you work through it..

tis true, i took it last semister and i cant remember nothing i learned

 

[Sukhoi]

Thanks for the help so far tonight guys. I'm busy studying for a big history test tomorrow, so no more time for math tonight. But tomorrow evening I'll post an example problem so everyone has an idea of what I'm talking about. 🙂

 

[Syringer]

Originally posted by: RaynorWolfcastle

Originally posted by: Syringer
Hmm, I'm not sure about v substitutions, but we've done u subs which is probably the same thing if you're still in the early stages.

Basically you have an equation that's in the form:

M(x,y) dx + N(x,y) dy = 0

All you have to do is do two substitutions:

let y = ux (or x = uy if it's easier to integrate the M function than N)

and let dy = x du + u dx (via product rule)

and just go ahead and plug it all in..simplify as much as possible, and then you should have a separable equation in terms of u and x, or u and y. Integrate it as usual, then plug y/x for all u's back in (or x/y). If you have a sample problem I can help you work through it..

you're describing an exact equation 😉

Well an exact equation has the form M(x,y) dx + N(x,y) dy = 0..but there is a difference in solving them. I forgot to explain it but you'd use the substitution when it's a homogenous equation..otherwise you use the method for exact equations..with partial derivatives and all that.

 

[RaynorWolfcastle]

Originally posted by: Syringer

Originally posted by: RaynorWolfcastle

Originally posted by: Syringer
Hmm, I'm not sure about v substitutions, but we've done u subs which is probably the same thing if you're still in the early stages.

Basically you have an equation that's in the form:

M(x,y) dx + N(x,y) dy = 0

All you have to do is do two substitutions:

let y = ux (or x = uy if it's easier to integrate the M function than N)

and let dy = x du + u dx (via product rule)

and just go ahead and plug it all in..simplify as much as possible, and then you should have a separable equation in terms of u and x, or u and y. Integrate it as usual, then plug y/x for all u's back in (or x/y). If you have a sample problem I can help you work through it..

you're describing an exact equation 😉

Well an exact equation has the form M(x,y) dx + N(x,y) dy = 0..but there is a difference in solving them. I forgot to explain it but you'd use the substitution when it's a homogenous equation..otherwise you use the method for exact equations..with partial derivatives and all that.

Either way, these are pretty short and sweet. Let me know when they teach you the series solution to higher degree polynomials (Frobenius method). You have no idea how many times I cursed out Frobenius and his ridiculously long and boring method last semester