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khalissa

  • 2 years ago

how to differentiate methods used in solving first-order differential equations?like separable,homogeneous equation,exact,etc.

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  1. vf321
    • 2 years ago
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    It's been a while since I took DE, but since separable is, well, separable, then it must ultimately be expressed as\[F(y)dy=G(x)dx\]which implies that it was originally\[\frac{dy}{dx}=\frac{G(x)}{F(y)}\]Thus, if you ever see a problem with \(\frac{dy}{dx}\) on one side and something which can group into two distinct functions for x and y that are all one term on the other, you know it's separable.

  2. vf321
    • 2 years ago
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    A homogenous function \(f(x, y)\) of degree \(n\) can be expressed as \[f(\alpha x, \alpha y)=\alpha^nf(x, y)\]When you encounter a DE, if it can be simplified into the following form:\[\frac{dy}{dx}=f(x, y)\]And then you check that \(f\) is homogenous. If it is, use the substitution of \(v=y/x\) and replace every instance of \(y\) with \(x v\). You'll notice the LHS becomes \(x\frac{dv}{dx}\). The DE will now be separable for v and x.

  3. vf321
    • 2 years ago
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    Also of note, a homogenous LINEAR DE implies that \[f(x)\frac{dy}{dx}+g(x)y=0\]The solution to this is just a special case of the solution to ALL first-order linear DEs, which can be re-written in the form \[\frac{dy}{dx}+f(x)y=g(x)\]for different functions \(f\) and \(g\). Wikipedia has a splendid, concise, understandable proof for this. I encourage you to go through the steps of the proof whenever you solve these instead of plugging into the equation Wikipedia gives you at the end for y(x). http://en.wikipedia.org/wiki/Linear_differential_equation#First_order_equation

  4. khalissa
    • 2 years ago
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    thanks!it do helps a lot =)

  5. vf321
    • 2 years ago
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    However, there are also exact first-order DEs, which solve the slightly more complicated \[a(x, y)y'(x)+b(x, y)=0\]which is often written as\[Mdy+Ndx=0\]where \(M=a(x, y)\) and \(N=b(x, y)\). The solution to this is exact if, using subscript notation to denote partials, \(M_x=N_y\). This is because of Green's Theorem, actually. If you'd like to to find out why, then make a new question regarding this and then type @vf321 so that I can answer it. But given that condition, the solution will be what you get when you find the integral:\[\int Mdx\cup\int N dy=c\]

  6. vf321
    • 2 years ago
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    Whoops! That equation should be \[Ndy+Mdx=0\]with \(M=b(x, y)\) and \(N=a(x, y)\). The integral is right, however.

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