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Video tags
Subtitles
Russian
00:00:02
easy mate in this video we are going to solve the00:00:04
partial differential equation of the00:00:07
square which is represented by00:00:10
this square speech operator which is also00:00:13
called the plasty not applied to a00:00:16
function or equal to zero this is the00:00:18
equation of the homogeneous square and00:00:21
we are going to solve it in this case for00:00:23
rectangular coordinates. Later00:00:25
we will see how it is solved, for example in00:00:27
polar coordinates. Before starting to00:00:30
solve this equation, I ask you to00:00:32
please support me by giving this video a like00:00:34
and subscribe to my channel and00:00:37
also follow me through my00:00:38
social networks facebook twitter00:00:40
instagram00:00:41
and twitch we are going to solve this equation00:00:44
specifically in the case of two00:00:47
dimensions we can then represent00:00:49
the equation in the following way in00:00:51
rectangular coordinates the second00:00:54
partial derivative of a with respect to00:01:12
In00:01:14
this video is when they give us00:01:16
certain conditions that00:01:19
the function must satisfy and we are going to solve00:01:21
this equation for example in the00:01:24
next one in the next region which00:01:27
is a square x goes from 0 to 3 and00:01:30
goes from 0 to 3 to understand00:01:32
We are going to draw this region here00:01:35
we have the Cartesian plane and what00:01:37
this region means is that x00:01:40
varies from 0 to 3 and would take from 000:01:43
to 3 forming this00:01:46
square then what we are looking for is a function00:01:49
of two variables xy will be defined within00:01:53
of this region that satisfies this00:01:56
partial differential equation and that00:01:59
also satisfies some conditions on00:02:02
the border of this region, which are these00:02:04
lines in red. In this video I am going to00:02:07
solve the simplest case, which is the00:02:09
following:00:02:10
we have conditions on three of the00:02:13
sides of the square that are equal to00:02:15
zero and the fourth side is equal to a00:02:18
function of x to understand these00:02:21
conditions how they are understood00:02:23
here geometrically simply notice00:02:25
that the first condition says at zero00:02:28
point00:02:29
equal to zero that means when x is00:02:31
equal to 0 it is worth 0 here and it varies from 000:02:36
to 3 so it is here from 0 to 300:02:39
so it refers to this side of the00:02:41
square00:02:42
here the function we are looking for00:02:44
must be zero now the following00:02:46
condition in three comma which means00:02:49
when x is worth 3x is worth 3 here while it00:02:53
varies from 0 to 3 then here00:02:55
varying ye from 0 to 3 then it00:02:58
refers to this side of the square here it00:03:01
must also be worth 0 then the next00:03:03
condition says00:03:04
in x 0 equal to 0 that means that it is00:03:08
worth 0 that is worth 0 here while x00:03:12
must vary from 0 to 3, then00:03:28
3 from here to here00:03:30
here in general it must be some00:03:33
function of x in this video I am going to00:03:36
do it with this function x for 3 - x but00:03:39
it can be any other00:03:41
continuous function the only thing that will change is00:03:44
when calculating the series of fourier00:03:47
we will see that in a moment more00:03:49
so this is how we can understand00:03:51
the conditions on the boundary for this00:03:54
problem and we are going to start to solve it00:03:56
we are looking for a function of two00:03:58
variables that satisfies these four00:04:00
conditions and that satisfies of course00:04:03
the equation of la plaza00:04:05
What we are going to do is apply the00:04:07
method of separation of variables, we are going00:04:10
to assume that the function of x comma that00:04:12
we are looking for can be separated as00:04:15
a product of functions such that one00:04:18
function only depends on x and the other00:04:21
function only depends on che So we have00:04:23
a product here and what we are going00:04:26
to do is substitute this function in the00:04:28
square equation for that00:04:30
we need to calculate these derivatives that are00:04:32
very easily calculated from00:04:34
this expression the second derivative of00:04:36
us with respect to x is going to be calculated00:04:38
simply differentiating this function00:04:41
twice here we have the second00:04:42
derivative of m with respect to x and the n of00:04:46
happens exactly the same because as00:04:47
we are doing partial derivative with00:04:49
respect to In00:04:54
the same way we calculate the00:04:56
second derivative of us with respect to00:04:58
and in this case the function that is derived00:05:01
is n from which the second00:05:03
derivative is calculated while the m of x which00:05:06
is now like a constant in this00:05:07
derivative happens exactly the same and now00:05:10
We substitute these00:05:12
in the equation leaving this00:05:15
sum here equal to zero now00:05:19
we are going to pass this term to the other00:05:20
negative side and then we pass this mx00:05:24
by dividing to the left side and this nd and we00:05:27
pass it by dividing to the right side in00:05:30
this way we manage to separate the00:05:32
variables in such a way that on the00:05:34
left side we have a function that00:05:36
only depends on x and on the00:05:38
right side we have a function that00:05:40
only depends on that and see that00:05:42
we are saying that both functions are00:05:44
equal the only way in which a00:05:46
function that only depends on x can be00:05:49
equal to one that only depends on y is that00:05:52
this function is the constant function00:05:54
which we are going to represent with lambda, which00:05:57
is how00:05:59
that constant is traditionally represented and now we have to00:06:01
analyze three cases for lambda that the00:06:04
band is zero that lambda is positive or00:06:06
lambda is negative let's start00:06:09
with the simplest case which is that lambda00:06:12
is equal to zero then the equation00:06:14
we have is this one here00:06:17
we are going to solve for m00:06:20
me vi prime of x between mx equal to00:06:23
zero this mx what is here by dividing00:06:26
we pass it by multiplying we are left with zero times00:06:28
m which is zero so we are left with this00:06:30
equation mb prime of x equal to zero it is00:06:33
a second order ordinary differential equation00:06:36
to solve it we00:06:38
simply have to integrate00:06:40
twice when integrating once we00:06:42
simply obtain a constant and when we00:06:45
integrate again we obtain the following: a00:06:47
constant times x plus another constant.00:06:51
These two constants a and b still have to be00:06:53
determined.00:07:03
conditions or in zero comma equal to00:07:05
zero or in three like yesterday equal to zero and00:07:08
remember that we said that v x comma was already00:07:11
separated as m from x by nd and then00:07:14
this and third comma that should be00:07:18
understood as m from 0 by n from she and that00:07:21
has to give us equal to 000:07:23
from there we deduce that m of 0 must be00:07:26
equal to 0 because n of y we would not want it to00:07:29
be 0 if n of jeff was00:07:31
identically 000:07:32
then we directly obtain that ud x00:07:35
comay is the trivial solution 0 which00:07:37
at the moment we do not want, so00:07:40
for this reason m of 0 must be zero in00:07:42
the same way00:07:44
here m of 3 must be 0 well let's00:07:47
substitute these conditions in the00:07:49
solution that we just obtained00:07:51
we start with m of 0 which00:07:53
simply means replace the 0 in the00:08:07
at once00:08:09
we substitute it and that must be equal to zero00:08:12
so this is the same as saying that00:08:14
3 is 0 or if we divide the 3 we are00:08:17
left with a being zero so both00:08:19
constants gave us that it is equal to zero00:08:21
so md x It will be equal to 0 x 0, that is,00:08:26
0, and then since mx is 0, when substituting00:08:29
here for00:08:39
condition that a00:08:43
x3 is equal to x times 3 - xs00:08:46
because if an00:09:00
case case number 200:09:03
lambda greater than 0 in this case it is convenient to00:09:05
represent lambda as a squared number00:09:08
since it is greater than 0 there is00:09:11
no problem with this lambda can00:09:13
be expressed as omega squared00:09:16
remember that when we square00:09:18
any real number the00:09:20
result is always greater than or equal to 000:09:22
so in this case00:09:24
we would simply discard the omega equal to 0 which would00:09:28
be the case that it gives us00:09:30
and the result as 0 and we are left00:09:32
only with omega different from 0 and00:09:34
squared the hunger00:09:35
will always be greater than 0 that's why we can do00:09:38
this well let's place it here and we are00:09:41
left with this equation00:09:43
again we are going to start solving for m00:09:45
so this m of x that is dividing00:09:48
happens by multiplying and we are left with this00:09:50
equation now we pass this by subtracting00:09:52
to the left side and This is an00:09:54
ordinary second-order differential equation00:09:56
with constant coefficients because00:09:59
omega is a constant. This equation is00:10:02
solved by means of the00:10:03
characteristic equation, which will be r squared minus00:10:06
omega squared. We calculate the two roots00:10:08
and we are left with a linear combination of00:10:10
exponentials like this one. I00:10:12
no longer do that here in detail because00:10:14
we already saw it in the00:10:16
ordinary differential equations course, but if you want to00:10:18
review that topic I am going to leave you in the00:10:20
description the link to the course where00:10:22
you can see how these types00:10:25
of equations are solved. We already have then the00:10:27
general solution for md x is a00:10:30
linear combination of this type of00:10:32
exponentials again we are going to substitute00:10:34
the initial conditions00:10:37
gm in 0 equal to 0 m / is equal to 0 well00:10:41
if00:10:44
steel is 1 so00:10:49
here it is directly c1 c2 equal to zero00:10:51
we have then an equation we are going to00:10:53
call it equation 100:10:56
now we are going to substitute x equal to 3 in00:10:59
this expression and we are left with this00:11:01
equation here that we are going to call00:11:03
equation 2 we are going to then solve00:11:05
this system of equations to00:11:07
find the value of c1 and c2 well from00:11:10
the first equation we are going to solve for 100:11:13
so we are left with that 1 is equal to00:11:14
less than two and now we substitute in00:11:17
equation 2 here in this s1 we put minus00:11:20
c 2 and now we can factor the c 200:11:23
leaving x minus raised 3 omega00:11:27
higher than minus 3 omega equal to 0 and00:11:29
then we have a multiplication00:11:31
equal to zero that means that either the00:11:35
first factor is 0 or the second factor00:11:37
is 0 but this second factor cannot00:11:41
be 0 the only way00:11:43
this factor could be 0 is if omega00:11:47
is 0 because then we would have less 1100:11:50
than 0 but as we said a moment ago00:11:53
we are going to assume that omega is different from00:11:56
0 because we want lambda to be00:11:57
strictly greater that 0 then since00:12:00
omega is different from 0 this factor is not00:12:02
zero and we are left directly with the fact that 200:12:04
is equal to 0 but we return to this00:12:08
expression above if 12 000:12:10
then one is equal to minus 0 or00:12:12
one becomes 0 then again the two00:12:15
constants are 0, reaching that mx is00:12:19
equal to zero, so x will already be 000:12:23
when substituting here and as we said a00:12:26
moment ago this does not satisfy one of the00:12:28
conditions it does not satisfy this condition00:12:31
here therefore We also discard00:12:34
case number 2 and move on to the case where00:12:37
we are missing case number 3. Lambda less00:12:39
than zero, just as before, it is convenient00:12:42
to express it with some00:12:43
squared quantity but now multiplied by a00:12:46
negative sign since omega square00:12:48
will always be a number,00:12:50
but when you multiply it Because of this00:12:52
negative sign it makes this always a00:12:54
negative number well let's put00:12:57
this here we obtain this00:13:00
equation and again we solve for m00:13:03
this mx happens by multiplying this term00:13:07
we pass it by adding to the left side and00:13:09
we are left with an00:13:10
ordinary differential equation again of second order00:13:13
of constant coefficients that in this00:13:15
case will give us a combination of00:13:17
sines and cosines in this way, then00:13:21
we are going to substitute the initial conditions here00:13:22
to see that in this case we are00:13:25
no longer going to obtain the trivial solution,00:13:27
the conditions are these: here m in 000:13:30
equal to 0 m in 3 equal to 000:13:32
so we start with the first one00:13:34
we substitute x equal to 0 here and we have00:13:38
this expression here we have to remember00:13:41
the cosine of 0 is worth 1 and that the sine of 0 is00:13:43
worth 0 so We remove this term00:13:46
and we are left with one by one equal to zero, that00:13:49
is, 1 is equal to00:13:52
zero.00:13:53
Now we substitute the second condition x00:13:56
equal to 3 in this expression, but since00:13:59
1 is equal to 0, it is no longer necessary to write the00:14:01
term it contains. the car not00:14:02
only write the term that00:14:05
has a sine then c2 becomes c2 times the sine00:14:07
of 3 omega equal to zero we then have00:14:10
a product equal to zero this00:14:13
means that either 20 or the sine of00:14:16
3 omega is 0 if 2 were 0 again00:14:20
we would arrive at the solution of00:14:33
Here is a simple00:14:35
trigonometric equation, you just have to00:14:37
remember that for the sine of a00:14:39
quantity to be equal to zero, that quantity00:14:42
must be an integer multiple of pi, that00:14:44
is, in this case, 3 omega must be00:14:47
equal to n times pi where n can be 123,00:14:52
etc. in fact it could also be 000:14:54
but if it were 0 omega would be zero and00:14:57
we are discarding that case00:14:59
here we do not put the zero so three00:15:02
omega is equal to n times pi this 3 that00:15:04
multiplies we pass it by dividing and00:15:07
we obtain that omega is equal to n times pi00:15:08
over 3 So we already have what00:15:12
the expression for omega in this00:15:14
equation must be for there to be a non-00:15:17
trivial solution, so what we are going to do00:15:20
now is place this that we just00:15:23
obtained here in the expression that00:15:25
we obtained from mx c 1 is worth 0, so the00:15:28
term with the cosine we are not going to write it anymore00:15:30
and you see that in reality this00:15:33
omega is an infinite set of00:15:36
possible values, a value for each00:15:38
integer so in reality00:15:41
we have an infinite set of00:15:43
solutions mx those solutions we are going to00:15:46
represent with the subscript n for00:15:49
refer to the fact that there is a solution00:15:50
for each n, so the solution m 1 will be00:15:55
when we take n equal to 1 and00:15:57
it will correspond to the sine of px over 3 and00:16:00
when n is equal to 2 we have another solution00:16:02
which is when we substitute here and it becomes00:16:04
sine of 200:16:05
over three etc., that is to say, we are00:16:07
representing with ms n the00:16:11
infinite set of solutions, one for each00:16:14
integer n iv, for each of these00:16:16
solutions we have a constant, so we are00:16:21
also going to represent each constant with a subscript, we put it as00:16:23
a sub n, and then we have the00:16:26
functions m sub n we are missing the00:16:28
functions n well for that now we take00:16:33
this part of the equation this n of it00:16:35
that is dividing we are going to multiply it00:16:38
and we are left with this here00:16:41
now this term that is negative goes to the00:16:44
other positive side and we obtain the00:16:46
following equation.00:16:47
This equation can give us a00:16:50
combination of exponentials as we saw00:16:52
in case number 2 for m, but in this00:16:55
case it is better for us to write it with00:16:57
hyperbolic functions, which is another00:16:59
way of writing this same solution.00:17:01
The solution with hyperbolic functions00:17:04
is the following nd ye equal to a00:17:07
constant times the hyperbolic cosine of00:17:09
omega ye plus another constant times the00:17:11
hyperbolic sine of omega that remember that00:17:14
the non-hyperbolic cost and the00:17:15
hyperbolic sine are defined as a00:17:17
linear combination of exponentials and00:17:19
as in this case the solutions of this00:17:22
equation are exponential any00:17:25
linear combination of this type of00:17:26
exponential is also a solution00:17:27
that is what makes it valid that the00:17:30
solution can be written as00:17:32
hyperbolic functions well here we are00:17:35
now going to substitute the conditions so00:17:38
that we have not yet written them00:17:40
here only those conditions are for00:17:42
x but we have a third condition that00:17:44
tells us that at00:17:56
what00:17:58
then n of 0 must be 000:18:01
so we substitute this condition here00:18:04
and we are left with n of 0 equal to c1 by00:18:07
consensus and public of 0 plus co2 for the00:18:09
sine and public of 0 equal to 0 and this is00:18:11
where we see why It is convenient to write it00:18:13
as hyperbolic sine and00:18:15
hyperbolic cosine because at 0 they have the00:18:18
same behavior as the00:18:20
usual trigonometric functions, the00:18:23
hyperbolic cosine at 0 is one and the00:18:25
hyperbolic sine at zero is zero, so00:18:27
we remove this term because it gives us00:18:29
zero and here we are left with one times one00:18:32
equal to zero, that is, if00:18:34
one is equal to zero, we already have00:18:37
the value of a constant, now here we have00:18:40
to be careful because00:18:41
we cannot calculate the other constant00:18:44
from the condition that we are missing00:18:46
because in this case The condition that we are00:18:48
missing gives us the same as a function of00:19:08
but a hyperbolic sine is not equal to00:19:10
a polynomial as much as we want00:19:12
so there we have to do another00:19:15
little trick that we will see below00:19:16
so until now the only thing we have00:19:18
obtained is that if one is worth zero so00:19:21
the function in Medellin will simply be a function00:19:23
of type hyperbolic sine of00:19:26
omega porsche but as we said omega is00:19:29
an infinite set of solutions00:19:33
so for each value of n we obtain a00:19:35
value of omega and therefore a value00:19:38
of this n of and then just as00:19:40
we did with m we are going to represent00:19:42
all the possible infinite set of n00:19:44
with a subscript we put n sub n of g00:19:49
equal to a constant that we are going to put00:19:52
as b sub n times the hyperbolic sine of00:19:55
omega why but omega is n pi over 300:19:59
so there is n piqué over 3 now00:20:02
We then have the functions in s n00:20:05
n00:20:06
and since it is a product of those functions00:20:08
we then obtain an infinite set00:20:11
of functions or sub n which will be to00:20:14
multiply the corresponding g emes vn00:20:17
by gene sub n leaving us with this00:20:20
product here00:20:21
in this product see that n are00:20:25
constants so by multiplying those00:20:27
two constants what we obtain as a00:20:29
result is another constant that we are going to00:20:31
better represent as c sub n this00:20:36
here is a solution for each value of00:20:39
n well then we have arrived at the00:20:43
following expression the solutions of the00:20:44
laplace equation are of this type,00:20:47
these solutions satisfy three of the00:20:50
boundary conditions, which are the ones00:20:52
that you see are equal to zero, but we00:20:54
need to satisfy the fourth condition.00:20:56
Here, what we are going to use is a00:20:59
property that tells us that any00:21:01
linear combination of these functions00:21:06
Any finite linear combination will also be a solution but00:21:09
also a linear and00:21:11
infinite combination that includes all these00:21:13
possible solutions and in fact that is the one00:21:15
that we will use in this case because00:21:16
only in that way will we achieve00:21:18
that the fourth condition is satisfied, that is,00:21:22
what What we are going to do is that00:21:23
our solution that we are looking for of00:21:37
What is00:21:40
missing is that it tells us that vd x 3 is equal to00:21:42
x times 3 - x in the interval 0 less than00:21:58
equal to this expression here, see00:22:01
that 3 is still valid, to replace here this 300:22:04
with this 3 below, they cancel and we are00:22:06
left with simply the hyperbolic sine of n,00:22:08
so we have that this sum must be00:22:11
equal to x times 3 x,00:22:13
well, in this expression, see that00:22:17
The only function that contains the00:22:19
variable x is the sine, the00:22:22
hyperbolic sine does not contain any variable00:22:36
n will be the product of this se his penis00:22:39
with this hyperbolic gentleman and by00:22:41
writing it this way we00:22:43
clearly see that what00:22:44
we are actually looking for are the coefficients b00:22:47
sub00:22:48
n of the fourier series in sines for00:22:51
the function x times 3 - x in the interval 000:22:55
less than00:23:09
a 200:23:12
over00:23:13
it is in this case 3 so it is 2 over00:23:16
3 of the integral from 0 to which is00:23:19
3 times the function x 3 - x x the sine of n times00:23:25
pi times x over l which is 3 well00:23:29
I have already done this explained in other videos00:23:30
previously about fourier series, I00:23:32
am going to leave you in the description the00:23:34
link to the playlist and I will00:23:36
also leave you the link00:23:37
directly to the video in which I calculate00:23:40
precisely this series of sines for00:23:42
this function exactly in that video00:23:45
we saw that The answer is to be here, we00:23:48
simply have to calculate this00:23:50
integral that is carried out by00:23:52
integration by parts00:23:54
twice, substitute the limits of00:23:56
integration and we will arrive at this00:23:59
result. Well, we already have the00:24:01
coefficients b sub n.00:24:03
Now with those coefficients we use this00:24:07
expression to obtain the coefficients.00:24:08
csn and be able to substitute it here and thus00:24:11
we will have the solution that we are00:24:12
looking for00:24:13
then sub n clearing from here00:24:17
this is the hyperbolic that is00:24:18
multiplying passes dividing we are00:24:20
then left with this division and here00:24:22
we substitute the expression b sub n which is00:24:25
to be here staying then all of this00:24:28
From here this is the value of the00:24:30
coefficients, it is assumed that we must00:24:33
substitute here, leaving us with this00:24:35
solution of x comma and e equal to this entire00:24:39
expression times the sine of n p x over 3 times00:24:42
the hyperbolic sine dnp times jr over00:24:44
3 this function of already satisfies the00:24:48
plant equation and the four00:24:50
boundary conditions, therefore,00:24:52
the solution we were looking for and so00:24:54
we have finished here. We could also00:24:57
simplify the expression a little by noting00:25:00
that we can separate in two cases if in00:25:03
that even or n is odd if in sparc this00:25:06
here gives us the result 1 and 1 - 100:25:09
gives us zero so all this becomes zero when n is even and00:25:12
when n is odd we express it in the00:25:14
form two beds one and then minus one00:25:18
raised to an odd exponent minus one00:25:19
times this less gives us more left one plus00:25:22
one which is 2 times 36 gives us 72 and then00:25:26
the coefficients will be like00:25:29
this and the solution of x as it is already00:25:32
written00:25:33
this way here we put the value00:25:36
of the coefficients and each n changes by00:25:39
two beds 1 in this way and the sum00:25:42
must start from here equal to 0 since00:25:44
the first odd number is generated for00:25:47
that value we are left with 2 x 0 + 1 which is 100:25:50
the first odd number so this is00:25:52
another way of writing the solution,00:25:54
either way is equally00:25:56
valid. I am infinitely grateful to the00:25:59
people who support me with their membership00:26:01
through YouTube and also through00:26:03
pechón, thank you very much to all of you00:26:06
here you can see a list with your00:26:09
namesDescription:
📩¿Necesitas ayuda con ejercicios? https://www.facebook.com/unsupportedbrowser 📲 . Anterior: https://www.youtube.com/watch?v=Y39CvSkSUNI Siguiente: https://www.youtube.com/watch?v=HFJ711jUtXI Ecuaciones Diferenciales Parciales EDP: https://www.youtube.com/playlist?list=PL9SnRnlzoyX05Y-DlDAoD4KwuHeNoP39F En este video veremos un ejemplo resuelto (ejercicio resuelto) de una ecuación en derivadas parciales de Laplace homogénea (ecuación diferencial parcial de potencial) en dos dimensiones definida sobre un rectángulo en coordenadas cartesianas rectangulares. Para esto usamos el método de separación de variables (Método de Fourier), factorizando la función de dos variables, derivando y sustituyendo en la ecuación y calculando a partir de eso los eigenvalores (valores característicos o valores propios) y las eigenfunciones (funciones características o funciones propias), analizando los tres posibles casos para lambda, resolviendo en senos y cosenos, y funciones hiperbólicas, y después usando Teorema de superposición de soluciones, para escribir la solución como una combinación lineal infinita de las funciones propias, y calculando la serie de Fourier en senos para el polinomio dado mediante integración por partes dos veces, obteniendo así el valor de los coeficientes y llegando finalmente al resultado como una serie. Todo explicado paso a paso de forma sencilla y fácil. __________________________________ ** ENLACES IMPORTANTES ** Ecuaciones Diferenciales Ordinarias EDO: https://www.youtube.com/playlist?list=PL9SnRnlzoyX0RE6_wcrTKaWj8cmQb3uO6 Ecuaciones Diferenciales Parciales EDP: https://www.youtube.com/playlist?list=PL9SnRnlzoyX05Y-DlDAoD4KwuHeNoP39F Álgebra Lineal: https://www.youtube.com/playlist?list=PL9SnRnlzoyX32lX7zNawatnGQP7IPLIi5 Variable Compleja: https://www.youtube.com/playlist?list=PL9SnRnlzoyX1EyKrhu12qtHyxrvAkLHHR Cálculo de Varias Variables: https://www.youtube.com/playlist?list=PL9SnRnlzoyX2-qH2lY3o5Lhv9f6za9o9A Integrales dobles y triples: https://www.youtube.com/playlist?list=PL9SnRnlzoyX07cHRqkJFoq6sPEfVPeIqT Videos Exclusivos: https://www.youtube.com/playlist?list=UUMOHwtud9tX_26eNKyZVoKfjA Curso de repaso de matemáticas (preuniversitarias) https://www.youtube.com/playlist?list=PL9SnRnlzoyX1-FFtFcUupLSdnTRvs8B5K __________________________________ ** MIRA TODOS MIS CURSOS AQUÍ ** https://matefacil.net/ __________________________________ ** BIBLIOGRAFÍA ** - Ecuaciones Diferenciales Parciales, Richard Haberman - Matemáticas Avanzadas para ingeniería, Peter V. O'Neil - Matemáticas Avanzadas para Ingeniería, Dennis Zill - Introduction to Partial Differential Equations, Peter J. Olver __________________________________ ** DONACIONES ** - Paypal: https://www.paypal.com/donate/ - Membresías del canal: https://www.youtube.com/channel/UCHwtud9tX_26eNKyZVoKfjA/join - Patreon: https://www.patreon.com/matefacil __________________________________ ** MIS OTROS CANALES Y REDES SOCIALES ** - Grupo de Telegram: https://t.me/matefacilgrupo - Canal de Física: https://www.youtube.com/channel/UCeFNpG-n8diSNszUAKaqM_A - Ejercicios de libros: https://www.youtube.com/@MateFacilEjercicios - Canal de Videojuegos: https://www.youtube.com/channel/UClSpw-rlRdygJmI33x1YagA - Twitch: https://www.twitch.tv/matefacil - Facebook (Página): https://www.facebook.com/unsupportedbrowser - Twitter: https://www.twitter.com/matefacilx - Instagram: https://www.facebook.com/unsupportedbrowser - TikTok: https://www.tiktok.com/@matefacilx - Discord: https://discord.com/invite/Gmb7sF9 __________________________________ #Matefacil #Matematicas #Math #tutorial #tutor #tutoriales #profesor __________________________________
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