A community for students.
Here's the question you clicked on:
 0 viewing
khalissa
 3 years ago
how to differentiate methods used in solving firstorder differential equations?like separable,homogeneous equation,exact,etc.
khalissa
 3 years ago
how to differentiate methods used in solving firstorder differential equations?like separable,homogeneous equation,exact,etc.

This Question is Closed

vf321
 3 years ago
Best ResponseYou've already chosen the best response.1It'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.

vf321
 3 years ago
Best ResponseYou've already chosen the best response.1A 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.

vf321
 3 years ago
Best ResponseYou've already chosen the best response.1Also 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

khalissa
 3 years ago
Best ResponseYou've already chosen the best response.0thanks!it do helps a lot =)

vf321
 3 years ago
Best ResponseYou've already chosen the best response.1However, 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:\[\int Mdx\cup\int N dy=c\]

vf321
 3 years ago
Best ResponseYou've already chosen the best response.1Whoops! That equation should be \[Ndy+Mdx=0\]with \(M=b(x, y)\) and \(N=a(x, y)\). The integral is right, however.
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.