#diff eq
168 messages · Page 1 of 1 (latest)
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Are you on High school mate?
So???
I am in classe préparatoire aux grandes écoles PCSI
I don't have same classes as you
...
I told you
Lemme guess...
same as highest level of calculus, algebra, and analytics etc for you
You're...in college?
That's good btw
In france we call that a preparatory class for highest level school
you know Ecole polytechnique or even ENS Ulm
You know mate...
Im still 9th grade 💀
But, after i looked around me...math is incredible! And fun!
Where did you get this question from?
What was the question you needed ?
euler identity ?
Yes
what don't you understand ?
Im still studying about that
not the answer I was looking for
Ik
what don't you get then
I forgot...ummm...
Ouh. The Taylor
?
I just don't understand about taylor series
Yup
it is a generalization of nth square root
for n> 1
and euler identity is when e =2
$e^x = \sum_{i = 0}^\infty \frac{x^n}{n!}$
ΛηοΣιμμυσ Ριδδληρ
it's not I dont think
it's not, the general solution has a log term
no
$\dv{y}{x}+\frac{1}{1+x}y=\frac{1+\ln(1+x)}{1+x}$, integrating factor is $\mu(x)=\exp(\int\frac{1}{1+x}\dd{x})=1+x$, so $\dv{x}[(1+x)y]=1+\ln(1+x)$
Omegabet_
hence $(1+x)y=\int 1+\ln(1+x)\dd{x}=x+(x+1)\ln(x+1)-x+C=(1+x)\ln(1+x)+C$
Omegabet_
so $y=\ln(1+x)+\frac{C}{1+x}$, look at that log term
Omegabet_
as for why you didnt get the log term in your general solution, you did something wrong.
@main kernel what is dy & dx means?
also that has nothing to do with Taylor series tbh, it's just Vieta's results applied to z^n-1
also I assume the radio silence on your part is you digesting the solution given
ProteinBuck
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oh you're doing variation of parameters?
so you have $y_p(x)=\frac{1}{1+x}\lambda(x)\to \lambda(x)=(1+x)\ln(1+x)$
Omegabet_
hence $y_p(x)=\ln(1+x)$ since $\frac{1+x}{1+x}=1$
Omegabet_
the particular solution isnt just lambda, it's (1/(1+x))lambda in this instance
yes
but so it is general + particular
hence lambda/1+x + (x+1)ln(1+x)
either it should be lmabda + ln(1+x)
i don't get why it is what you said
even tho it is true
can you call to explain ?
recall how variation of parameters works
ok
you find the homogeneous solution, which was $y_h=\frac{C}{1+x}$
Omegabet_
yes
then you guess that the particular solution is of the form $y_p=y_h\lambda$
Omegabet_
to which the general solution is $y=y_h+y_p$
Omegabet_
you found $\lambda=(1+x)\ln(1+x)$ and $y_h=\frac{1}{1+x}$
Omegabet_
so what's $y_p$?
Omegabet_
maths lore is cool, maths gameplay just sucks
let me do it all over again for the third time
third time's the charm
you didnt do anything wrong in your working, you just plugged in lambda for your particular solution, not y_h*lambda
yeah, so what's the particular solution?
there you go
How bad am I bruh
and no lol, making a mistake doesnt mean you stop forever
Good redaction ?
Look how dumb the mistake is
so?
How can i pretend at high level failing at this
cause higher level doesnt really do computations
computation ?
yeah, solving ODEs is computation
or for computers
pre hilbert spaces specifically seems like a weird niche but ok
that's just a vector space w/ an inner product, you usually care about hilbert spaces in which the induced norm makes the space complete
ok yeah, just defining the notion of inner products
I am still at really low level maths
I can’t afford doing mistakes like that that’s just why am disappointed
anyway, going back to that physics shit

then you just take your time until the algorithms/steps are crystal 
your mistake was just forgetting how the particular solution was constructed, the actual work was all correct
what is your maths level

