#Need help with Nonhomogeneous Differentail Equation!
58 messages · Page 1 of 1 (latest)
using undetermined coefficients you would guess an appropriate solution and solve for the required coefficients
sorry, I'm not really following 🙁
how have you dealt with non-homogeneous equations before?
so we are supposed to make this in the form y non-homo = y homo + y s, where s means 'special'
I'm sorry, I'm a native German and not used to the English technical words
are willing to guide me through my dispair? 😅
yes but have you come up with the particular solution before? if so, how?
thank you! no, I haven't and that's where I'm stuck currently
I tried to differenciate y homo and tried to insert it in my original equation, but this didn't quite work out
well have you come up with particular solutions for this type of equations before?
no not really 😓
and I dont really understand how to obtain it
I know you're somewhere supposed differentiate it and insert it somewhere, right?
so the idea for this method that we will guess a solution which has some unknown constants in it, and then plug it into the equation
this page has lots of examples of how to do that:
https://tutorial.math.lamar.edu/Classes/DE/UndeterminedCoefficients.aspx
In this section we introduce the method of undetermined coefficients to find particular solutions to nonhomogeneous differential equation. We work a wide variety of examples illustrating the many guidelines for making the initial guess of the form of the particular solution that is needed for the method.
so that really the farthest i have come
thank you! I will look into it
can I come back to you if I still dont understand it? 😅
sure
thx
you have the right idea for the guess, but you run into the problem that it is part of your homogeneous solution
if you have that situation, you have to multiply it by the x^s, where s is the number of times e^x showed up as part of the homogenous solution
why ^3?
sorry!
you said as much as e^x shows up in my homo solution, which would be 3 times
and what do I multiply x^s with?
your homogeneous solution has e^x twice and e^-x once
i think you maybe forgot to copy the - sign in your notes here
yeah, sorry!
what about that?
hold on! I found the example on your website where this is discussed XD
let me read this real quick
is that correct? it took me almost an hour to write all of that down 😭
pls say it's correct 👉 👈
it appears to be correct
holy fucking shit
tyyyy
but I have one last question
why do we guess with (Ax+B) in the case of xe^x / a doubling in solutions, when we subsequently can stick with a simple A and B?
i'm not sure what you mean by "a simple A and B"
in a more simple OED, where the right side is i.e. a Polynomial, you can stick with simple variables like A * p(x) + B * q(x) +..., yet in a non-linear right side like xe^x, you have to use (Ax+B), or am I wrong?
I'm sorry if I mess up some technical terms, I'm fairly new to this topic - you have to get the gist though XD
if the right side is a polynomial, your guess is a polynomial of the same degree
e.g. right side is x^2 + 3, you guess Ax^2 + Bx + C
if that polynomial is multiplied by e^x you multiply your guess by e^x to match
so why did we guess Ax+B for xe^x ?
because x is a first order polynomial
like if the right side was just x we would guess Ax + B, all that changed is that it was multiplied by e^x
sorry, I still don't really understand your derivation
It might be due to the late hour from where I live
I really have to thank you for your help
I could never have done it without you - so thank you!