Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

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

Mathematics
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Join Brainly to access

this expert answer

SEE EXPERT ANSWER

To see the expert answer you'll need to create a free account at Brainly

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.
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.
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

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

thanks!it do helps a lot =)
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:\[\int Mdx\cup\int N dy=c\]
Whoops! That equation should be \[Ndy+Mdx=0\]with \(M=b(x, y)\) and \(N=a(x, y)\). The integral is right, however.

Not the answer you are looking for?

Search for more explanations.

Ask your own question