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Ordinary Differential Equations (ODE)

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Hey all. I took Linear Algebra last fall, but the ODE stuff was never cemented in me, nor did I practice enough. In fact, I forgot how to solve them by the middle of Spring 2017 semester.

So, before school starts back up in 2 weeks I want to practice my ODE's again. I picked up a practice book by Schaum (pic related) which is awesome. Tons of solved problems in case I get stuck, and quick introductions to each type of ODE.


My question however, is more on the "why". What exactly is a first order homogenous equation, and why does making the substitution of [math]y=vx[/math] or [math]v= \frac{y}{x}[/math] work?

Additionally,

For Exact first order differential equations, what is all this about [math]M(x,y)[/math] and [math]N(x,y)[/math]? I can solve these problems easily, but what's the deeper meaning behind these solutions and why do they work?
>>
>>9123481
>homogeneous
Read the definition. Suppose that we have [math] f,g [/math] such that they are nth degree homogeneous and we have the differential equation

[math] f(x,y)dx + g(x,y)dy =0 [/math].

So consider the substitution y=ux, so dy = udx + xdu. We then get:
[math] f(x,ux)dx + g(x,ux)(udx + xdu) =0 [/math].

By the definition of nth degree homogeneous, this turns into
[math] x^nf(1,u)dx + x^ng(1,u)(udx + xdu) =0 [/math]. And you notice there is something you can cancel.

[math] f(1,u)dx + g(1,u)(udx + xdu) =0 [/math].

And now just rewrite it in a cute way.
[math] g(1,u)xdu = (f(1,u) - g(1,u)u)dx[/math].

And now even cuter.
[math] \frac{dx}{x} = \frac{g(1,u)du}{f(1,u) - g(1,u)u}[/math].

The left hand side is a function of x. The right hand side is a function of u. It just werks because it is separable. Maybe I did the algebra wrong but you get the idea. Homogeneous = we can cancel out x almost completely and get rid of all the complexity of the equation.

Gonna do a follow up on exact ones. Give me a min.
>>
>>9123481
>Exact.
Read the definition.

Anyways, exacts are now less about algebra and more about analysis. If you did calculus 3 then you know where this comes from and if not then I guess it is time you study vector fields. Anyways, even if you do not understand what I am about to say then at least I will be pointing you into the right direction.

Anyways, you know that if [math] F [/math] is a vector field then

[math]dF = F_x dx + F_y dy [/math]

And you also know that if F is a vector field then setting [math] F = C [/math] generates an implicit function in a dimension 1 smaller than the dimension F is defined in. But if [math] F = C [/math] then [math] dF = 0 [/math] which implies [math] F_x dx + F_y dy = 0 [/math]

So F as a generator of implicit functions is a solution to the differential equation it itself generates. However when you are given problems you are not given F. You are given M and N, but you hope in your heart of hearts that there exists some F such that [math] F_x = M, F_y = N [/math] because if that is the case then your differential equation is just the differential equation generated by some vector field so now all you need to do is find the vector field and then find the family of functions the vector field generates when it is "converted" to an implicit function.

Anyways, Calc 3 gives a sort-of rigorous treatment of implicit functions in general dimensions, vector fields and differentials so to understand this better just research those topics.
>>
>>9123566
Nice
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