#Need help with Nonhomogeneous Differentail Equation!

58 messages · Page 1 of 1 (latest)

sacred fractal
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Hey, I’m currently studying for my upcoming Maths for Engineers III exam and need help with that math problem:

Find the solution for that nonhomogeneous differential equation: y‘‘‘-y‘‘-y‘+y = xe^x.
I‘ve already found the homogeneous solution y homo = c1e^x + c2xe^x + c3e^-x. How do I continue?

foggy pendantBOT
steel holly
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using undetermined coefficients you would guess an appropriate solution and solve for the required coefficients

sacred fractal
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sorry, I'm not really following 🙁

steel holly
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how have you dealt with non-homogeneous equations before?

sacred fractal
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so we are supposed to make this in the form y non-homo = y homo + y s, where s means 'special'

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I'm sorry, I'm a native German and not used to the English technical words

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are willing to guide me through my dispair? 😅

steel holly
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yes but have you come up with the particular solution before? if so, how?

sacred fractal
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thank you! no, I haven't and that's where I'm stuck currently

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I tried to differenciate y homo and tried to insert it in my original equation, but this didn't quite work out

steel holly
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well have you come up with particular solutions for this type of equations before?

sacred fractal
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no not really 😓

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and I dont really understand how to obtain it

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I know you're somewhere supposed differentiate it and insert it somewhere, right?

steel holly
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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

sacred fractal
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so that really the farthest i have come

sacred fractal
steel holly
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sure

sacred fractal
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thx

sacred fractal
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so uhh

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I'm kinda stuck again...

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did I do smt wrong? 😢

steel holly
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you have the right idea for the guess, but you run into the problem that it is part of your homogeneous solution

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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

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why ^3?

sacred fractal
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sorry!

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you said as much as e^x shows up in my homo solution, which would be 3 times

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and what do I multiply x^s with?

steel holly
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your homogeneous solution has e^x twice and e^-x once

steel holly
sacred fractal
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yeah, sorry!

sacred fractal
steel holly
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your guess for y_p

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so now it is x^s e^x (Ax + B)

sacred fractal
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hold on! I found the example on your website where this is discussed XD

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let me read this real quick

sacred fractal
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is that correct? it took me almost an hour to write all of that down 😭

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pls say it's correct 👉 👈

steel holly
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it appears to be correct

sacred fractal
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holy fucking shit

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tyyyy

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but I have one last question

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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?

steel holly
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i'm not sure what you mean by "a simple A and B"

sacred fractal
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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?

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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

steel holly
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if the right side is a polynomial, your guess is a polynomial of the same degree

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e.g. right side is x^2 + 3, you guess Ax^2 + Bx + C

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if that polynomial is multiplied by e^x you multiply your guess by e^x to match

sacred fractal
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so why did we guess Ax+B for xe^x ?

steel holly
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because x is a first order polynomial

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like if the right side was just x we would guess Ax + B, all that changed is that it was multiplied by e^x

sacred fractal
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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!