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khalissa
Group Title
how to differentiate methods used in solving firstorder differential equations?like separable,homogeneous equation,exact,etc.
 one year ago
 one year ago
khalissa Group Title
how to differentiate methods used in solving firstorder differential equations?like separable,homogeneous equation,exact,etc.
 one year ago
 one year ago

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vf321 Group TitleBest ResponseYou've already chosen the best response.1
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.
 one year ago

vf321 Group TitleBest ResponseYou've already chosen the best response.1
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.
 one year ago

vf321 Group TitleBest ResponseYou've already chosen the best response.1
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 firstorder linear DEs, which can be rewritten 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
 one year ago

khalissa Group TitleBest ResponseYou've already chosen the best response.0
thanks!it do helps a lot =)
 one year ago

vf321 Group TitleBest ResponseYou've already chosen the best response.1
However, there are also exact firstorder 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:\[\int Mdx\cup\int N dy=c\]
 one year ago

vf321 Group TitleBest ResponseYou've already chosen the best response.1
Whoops! That equation should be \[Ndy+Mdx=0\]with \(M=b(x, y)\) and \(N=a(x, y)\). The integral is right, however.
 one year ago
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