Up to my knowledge, there are mainly 3 ways of solving these equations: I) undetermined coefficient method II) D-operator method III) variation of parameter
Do you guys know any resources or books that can explain the operator D method ? I don't seem to find this method in any book I have read so far.
Hi,
I spent the weekend trying to solve this equation, but I can't find a way to get the correct solution.
To be honest, I have no idea how.
I would like to explain what I did, but I'm so confuse that it makes no sense.
I tried to solve as a first order equation of V then integrate once more to have an equation for z.
Any help would be appreciate.
the solution is z = z_0 + Tv_0[1- e^(-t/T)]
Where T = m/k and v(t=0) = v_0
So I suppose I'm asking 2 questions.
The first one is the "fundamental set of solutions". My book basically said that if y1 and y2 are solutions to the DE: y'' + py' + qy = 0, and wronskian of y1 and y2 is not 0, then y1 and y2 form a fundamental set of solutions. But what is meant by fundamental set? Is that saying that those are the only two solutions to the DE? Is the combination y1+y2 also within the set (I'm thinking back to span in Linear Algebra here ,so I'm not sure if that's correct)?
And there's another part of the book that states a theorem:
> Consider the equation: L[y] = y'' + p(t)y' + q(t)y = 0, whose coefficients p and q are continuous on some open interval I. Choose some point t0 in interval I.
> Let y1 be the solution of the equation: L[y] = y'' + p(t) y'+ q(t)y = 0 that also satisfies the conditions y(t0) = 1, y'(t0) = 0.
> Let y2 be the solution of the equation: L[y] = y'' + p(t) y'+ q(t)y = 0 that also satisfies the conditions y(t0) = 0, y'(t0) = 1.
> Then, y1 and y2 form a fundamental set of solutions.
...I'm confused as to what this is even saying? Why are we picking these arbitrary values 1 and 0? Why not pick some other value 2 or pi or 10? Why did they come up with 1,0 and 0,1 vs say 5,10 and 10, 5? The book then goes to prove that cosh(t) and sinh(t) is a fundamental set to the DE y'' - y = 0.
So... I'm getting super confused by this chapter. Solving the characteristic equation is easy, but this chapter completely blindsided the flow and I'm confused as to what these are even saying.
Any advice?
Lost. In a linear algebra class and fell behind... I have the Leon text. My next project is to "Use the matrix methods to find the general solution of each of the following homogeneous differential equations." Can anyone point me to a resource where I can figure this out?
Example problems:
x′=Ax, where A is a 2×2 matrix with rows 5 -1, 3 1
x′=Bx, where B is a 2×2 matrix with rows 2 3, -1 -2
Where are the shitposts? The nonlinear PDE erotica?
I’m leaving.
SSS out.
I took difs some years back but I’ve been reviewing them. It was never really explained to me (or at least not in a way that stuck) why the complementary solution needs to be added to the particular solution to get the general solution when solving a non homogeneous dif eq. If the particular solution satisfies the dif eq, why do we need to add the complementary solution to have the general solution? Why can’t the particular solution be the general solution? Sorry if this seems like a naive question but I just want to understand why. Thanks!
Hey all, I'm struggling with this problem and I'm not really sure how to go about it. Any help is appreciated! https://gyazo.com/3bc5d7174258ad245b6539f01e578c75
Solve the first order homogeneous differential equation x^(2)(dy/dx) = xy - y^(2) by making the substitution y = vx.
Some sort of working out would be appreciated bc I honestly have no idea how to do this. Thanks
New video Added to the Differential Equation Playlist. In this video, The step by step process of solving First Order Homogeneous Differential Equations with the substitution method is explained in a way that's comprehensive and very easy to understand.
I just spend two damn hours trying to calculate this and my result is nothing near the key result. Here's what I did:
- Isolate y' --> y' = - y / (y - x)
- Substitute 1 / (y - x) for v therefor y = 1/v + x therefor y' = (-1/u^(2))(du/dx) + 1
- Use substitutes and rearrange to -1/u^(3) du = x dx
- Integrate and simplify into y = 2x + c
Honestly, I feel totally lost. I feel like I must've done plethora of errors along the way but don't even ask me "how did you get from [this] to [that]" because I don't even freaking know anymore. I feel so stupid. Anyone can explain this so that even a smol brain like me can understand?
I have been struggling with this question all day...
What homogeneous linear differential equation has e^(x^2) and logx as solutions. (The solution should be of order of 2 or greater)
It seems to me that the answer should be an annihilator for e^(x^2) and logx but I can't find an annihilator that works.
Why cant we just take the particular solution and say that is the answer?
Hello, I need help doing this homogeneous differential equation second order:
I know this is also Euler's differential equation but I need to do it using using classic substitution y = e^(integral of zdx). When i apply the substitution I get really weird expression that I can't solve.
Here's the higher order linear differential equation:
a_0(x)y^(n) +a_1(x)y^(n-1) + ... +a_(n-1)(x)y^(1) +a_n(x)y +b(x)=0
All sources I've seen so far say when b(x)=0, the DE is homogeneous. I don't understand how it is homogeneous and haven't been able to find any material explaining the same.
The variable coefficients can be anything... that's what is messing with me.
Edit: tried fixing the formatting of the equation but cant figure it out. I guess it gets its point across in its current form
Is there a quicker way to find the form of the particular solution to a given 2nd Order NH DE? For example, for y”-3y’-4y=-8e^(t)cos(2t) the particular solution would be Y(t)= Ae^(t)cos(2t) + Be^(t)sin(2t). However, using the same form for y”+2y’+5y=12e^(-t)cos(2t) would not work. You find this out after doing all the busy work deriving. You then multiply it by t and try again. Is there a method/shortcut to determine the best particular solution form without having to do all this tedious work? I.e. Is there a way for me to know to multiply by t at the beginning without the trial and error?
Hi all!
I'm currently studying some methods for solving ODE's by myself. I am, unfortunately stuck on a definition.
According to the notes I'm reading, two equivalent definitions for homogeneous equations are that:
-
for an ODE P(x,y) dx + Q(x,y) dy = 0, the equation is homogeneous if P(kx,ky)/P(x,y) = Q(kx,ky)/Q(x,y)
-
An ODE is homogeneous if it is of form dy/dx = f(y/x)
Now, the claim that the ODE dy/dx = a(x) * y is homogeneous. I just can't see how that fits any of the definitions given :(.
All help is greatly appreciated!
TL;DR How is dy/dx = a(x) *y homogeneous?
y' + 2y + 1 = 2xy
I tried writing it as y' + (2 - 2x)y = -1 my opinion is that its not homogenous. My friend says that it is, because f(y',y'',...) = 0 is homogenous and you can just write
y' + 2y - 2xy + 1 = 0
Can someone elaborate how you find out whether a differential equation is homogenous?
Can someone seriously explain to me/give me a sense of intuition about this differential equation and its applications? I can solve the problems etc but can't seem to have an exact intuition...
Links/Sources and graphical explanations are much appreciated...
The problem in question is 3.17 a) from this book.
I have tried and failed at least 4 times now on what should be a straightforward question.
If anyone could lead me through at least one of the derivations of say (d^2 u)/(dx^2) in terms of (du)/ds and du/dt that would probably help me solve the problem because as is I am probably doing something wrong.
