PDP | PART 1

Assalamualaikum Wr Wb In this video we will learn about partial differential equations We start with the precedents PDP, namely partial differential equations that we will learn today, plays an important role in the visualization of the physical situation where the quantities involved in it change with respect to time and space

If the variables are related to the space, then the variable x, y will appear, if it is 2 dimensions, or z, if it is 3 dimensions. And if it changes to time, then there will be a variable t that describes the time.

For example, if we look at the topics of physical or advanced physics such as the classical mechanics of electromagnetic, hydrodynamic, and quantum mechanics, namely Schrodinger's waves, we will find

the use of partial differential equations which are used to describe the physical phenomena related to these problems. So here, first we will learn about the definition of the common differential equations and partial differential equations. We start with the differential equations which are often abbreviated as PD or

differential equation, which is abbreviated as DE. This differential equation is a equation that fits one or more decreases from an unknown function to one or more free variables. So here I draw thick, this is what we have to draw together, that in the differential equation,

The equation will be one or more decreases from an unknown function. Here, it is one or more free variables. So the number of

the descent of an unknown function is 1 or more and the free variable can be 1 or more so here the differential equation will be divided into two the first is the partial differential equation or usually abbreviated as PDB or commonly called as ordinary differential equation or OD

Here, the usual differential equation is a equation that fits one or more derivatives of an unknown function to one free variable. So what we need to pay attention to here is the number of free variables is only one. Meanwhile here, the second division of the differential equation is the partial differential equation or often abbreviated as PDP.

or often called Partial Differential Equation or PDE

This is a combination that contains one or more derivatives from an unknown function to two or more free variables. So we pay attention here that the number of free variables is two or more. So we understand the difference between the normal and partial differential combinations. Here, if the normal differential combination, the number of free variables is only one,

While in partial differential equations, the number of free variables can be 2 or more. Next, after we learn about the definition of differential equations, ordinary differential equations, and partial differential equations, then we must be able to classify the following examples, which are ordinary differential equations or partial differential equations.

We see in the first equation, here y' + xy = e^x. We know that y' = dy / dx, so y is reduced to the variable of x.

y is a bound variable or the independent variable while x is a free variable. We see here y' + xy = e^x is

the usual differential equation. Why? Because we pay attention here, the number of free variables involved in the partial derivative is only one, namely x. So, we can write the first equation here dy / dx + x * y = e^x.

It can be seen here that the function that is not known is y and the free variable is x. So here the first equation is called the usual differential equation because the equation fits

partial descent, namely dy per dx, where the associated variable is y and the number of the free variable is only one, namely x. Let's look at the second example here, y double accent square plus x equals y accent third. This is also a common differential equation because we pay attention here

y double accent is equal to d squared y per dx squared this is squared plus x equal to

dy/dx^3. It can be seen here that the number of the associated variable is 1, which is y, and the number of the free variable is also 1, which is x. So here the equation fits one free variable, which is x,

so it includes the usual differential equation. Let's look at the third example, here there is y double accent plus y accent minus 3y equals 0. This is also an example of the usual differential equation because the bound variable is y, here y is reduced

to one free variable, which is x. For the fourth example here, d^2/dt^2 - 3d^2/dt + 2^2 = sin2t. This is also an example of PDB.

because we pay attention here, the bound variable, which is the key, is reduced to one free variable, which is t. So here, the fourth equation is an example of the usual differential equation. Let's look at the fifth example, here dou squared z divided by dou x squared is minus dou squared z per dou y squared equals zero.

This is an example of partial differential equivalence. Why? We pay attention here, there is a bound variable, namely Z here. Z is reduced to two free variables, namely X and Y. So the number of free variables involved here in the partial derivative is 2, namely X and Y.

So if there is a similarity that satisfies the partial descent of a function that is unknown, where the number of free variables that are involved in it, two or more, is a partial differential similarity. So it is clear here, the fifth example is a partial differential similarity.

Next, for example 6, here there is a combination that matches the partial descent of a function that is not known, namely u. Here, u is reduced to two free variables, namely x and y. So the number of free variables involved in the combination of the 6 is two, namely x and y.

So it is clear that the 6th equation is also a partial differential equation. Next, we will discuss the order of a partial differential equation. We see here that a partial differential equation will have an order n if the descent to n is the highest descent in the partial differential equation.

For the first equation, we see that the highest descent in the partial differential equation

is the second derivative, here is dou squared u per dou x squared and here is also dou squared u per dou y squared. So the highest derivative is 2, so we can say that the partial differential equation is a partial differential equation of order 2, because the equation makes up the highest derivative, which is 2.

Next, for the second example here, dou squared u divided by dou x squared plus 5 dou u divided by dou y equals e to the x-th. We pay attention to this, there are two partial decreases, namely the partial decrease here, the first one.

The second is this. The first one, dou squared u per dou x squared, is the second derivative. The second derivative of u to the free variable x. While the second one here is the first derivative

from variable u to variable y, so the highest descent is the second descent here. So we can say that the partial differential equation is a partial differential equation with order 2.

Next, for the third example, we will see that dou of 3z divided by dou x dou y squared is minus 2xz multiplied by dou of z squared divided by dou y squared divided by 4 is equal to 0.

we see here there are two partial decreases for the first one here the third decrease from the variable related z while the second one is the second decrease so

The PDP is PDP order 3 because the highest decrease in the partial differential equation is the third decrease. Let's see the fourth example here. 3 times dou squared z per dou y squared squared plus dou squared z per dou x squared dou y equals 0. This is a partial differential equation that is order 3.

because we note here there are two partial decreases, namely the first partial decrease is the second decrease from the variable z and the second is the third decrease from the variable z so here the highest decrease is the third decrease so here we can say that the PDP is a PDP with order 3

Next, after we learn about the order of a partial differential equation, then we will continue the discussion about the degree of a partial differential equation. A partial differential equation will be of degree K if the highest derivative in the PDP has the highest level K.

Let's see the example so that it is easier for us to understand. Here is the partial differential equation, which is: dou squared u per dou x squared plus 2xy dou squared u per dou y squared plus u equals 1.

we have discussed that the partial differential equation is PDP with order 2 because the highest descent is level 2 while here we note that the degree of the PDP is degree 1 why? because the highest descent that fits in the PDP is the second descent has the highest level

So here if we pay attention that the highest drop has the highest level 1, this is the same level 1, so it is clear here that the first PDP is a level 1 PDP. Let's look at the second example, this is a PDP with order 2 as we have discussed before, because

the highest drop from the PDP is the drop of the fruit level, so here the order is order 2, while we pay attention that the degree is degree 1. Why? Because we see that

The highest descent, which is descent 2, has rank 1, so it is clear that the second PDP is PDP order 2 with degree 1. We look at the third example, here we have discussed that the PDP

is a PDP with order 3 because the highest drop in the equation is the third level drop. We note that the highest drop, the third level, is level 1. So this is level 1. So we can say that the PDP is level 1. So why not 4 here?

Because the fourth level is the second-level partial descent, while the second-level partial descent is not the highest descent in the differential partial equation. The highest is the third descent. So here we see the level of the highest partial descent.

which is 1, so as I said here, the third equation is PDP with an order of 3 degrees 1.

We see the fourth example here we have discussed before that the PDP is a PDP of order 3 because the highest partial drop in the PDP is a drop of level 3 while we see that the rank of the highest drop here is rank 1 so that we actually already see the partial drop

The highest of the sum is the third derivative and the sum is equal to the highest derivative which is rank 1. So we can say that the PDP is a degree 1 PDP.

Next, after we discuss the order and degree of a partial differential equation, then we will discuss the linearity of a partial differential equation.

We see here that a partial differential equation will be linear if the first function is a function that is unknown in a partial differential equation and the partial derivatives are in the first class or the second here, the coefficient of the temperature in PDP is only dependent on the free variable or constant

And if not, then the partial differential equation is a non-linear partial differential equation. Let's see here, after we know

about linearity of PDP, then we must be able to determine whether a partial differential equation is a linear or nonlinear partial differential equation. We see the first one here, dou squared u per dou x squared plus 2xy dou squared u per dou y squared plus u equals 1.

We have discussed before that the PDPA is ordered 2 because the highest drop in the partial drop is

the second descent and the degree is 1 because the highest partial descent has a rank of 1 and we see that every function that is not known or the partial descent is ranked 1 we pay attention here this is the partial descent from u is ranked 1 or here the attribute that is the associated variable is u this is also ranked 1

while the second we note that each class depends on the variable free or constant we see here this is the coefficient is constant, namely 1 here also the coefficient is 2xy where x and y are free variables and here the u

also the coefficient is 1, then here we can say that this first partial differential equation is a linear partial differential equation, because every

the function that is unknown or the partial descent is in the first class, there is no element that depends on the associated variable, so it is clear here in the first equation, this is a linear partial differential equation. For the second one here, dou squared u per dou x squared plus 5 dou u per dou y, this is equal to e to the power of x,

This is PDP order 2 because the highest partial descent has rank 2 and we note that the highest partial descent is rank 1 so here is a degree 1 PDP. We note here, every function that is unknown, namely u or the partial descent

is in the first rank, so this is in the first rank, then the second one here is also in the first rank. The second one here, each attribute depends on a constant or a free variable. Here is a constant, this is also a constant, so we can say that the second PDP is a linear PDP.

Third, we note that the PDP is an orderly and grade 1 PDP because the highest drop from the balance is a partial drop of level 3 and the highest partial drop has a level 1 so it is called a grade 1 PDP.

We note that the coefficient of the second element, which is here, -2xz times dou squared z per dou y squared square 4, has a coefficient that contains or fits the variable related to z. And here is this double.

In addition to the class that depends on the variable related to z, there is a partial descent that is not one, namely a fourth, so clearly the PDP is a non-linear PDP.

Next, for the fourth equation, here 3 times dou squared z per dou y squared is squared plus dou of the third level z per dou x squared dou y equals 0. We have previously discussed that the PDP is a PDP with order 3 because the partial descent contained in the equation is the third descent.

and because the highest partial descent has rank 1, then here the PDP is a degree 1 PDP. We note that here there is a partial descent that has rank 2.

So here we can say that PDP in the four equations is non-linear PDP because there is a partial descent that has a rank of more than one.

Next, here we see the fifth example. dou squared z times dou y squared square 3 plus 3 times dou squared 3 z times dou x times dou y squared equals x. We note that in the partial differential equation, the partial descent is

is the third highest partial decline, namely the highest partial decline in

the partial differential equation so we can say that the partial differential equation has order D3 we note that the partial descent with the highest level is 1 so the degree of PDP 5 is degree 1 next here in the partial differential equation 5 this makes the partial descent

is more than 1, which is 3 Therefore, we can say that PDP 5 is non-linear PDP

After we discuss the linearity of a partial differential equation, then we will discuss the homogeneous and non-homogeneous partial differential equations. A partial differential equation is said to be a homogeneous partial differential equation if all the tribes will fit an unknown function or partial descent.

If not, then the PDP is called as non-homogeneous PDP. Let's see the example. For this first comparison, as we discussed before, that the PDP is a PDP that has an order of two, one degree, linear, and we note that this is a non-homogeneous PDP because there is a non-suitable class,

a function or partial descent so here there is a class, namely here 1, 1 does not fit the unknown function, namely u or partial descent from u so here the first PDP is a non-homogeneous PDP for the next one, here the second PDP is also a non-homogeneous PDP

because here we pay attention to the first class, this is the partial descent of the U-threaded variable, here the second class also has a partial descent, but here there is a class, namely E-x, which does not make up the unknown variable, namely U or the partial descent, so here we can say that the PDP is a non-homogeneous PDP.

For the third example, we note that all tribes have a function that is unknown, namely z or its partial descent. This is the first tribe, this is the third-level partial descent. Then the second tribe, this also has a second-level partial descent.

which also depends on the variable z here so here we can say that this third PDP is a homogeneous PDP. Next, here for the fourth PDP, this is also a homogeneous PDP because each tribe fits

a function that is not known Z or the partial descent if here the fourth is the partial descent so here we can say that the PDP is the homogenous PDP for the fifth PDP we pay attention here there is a subordinate that does not fit

an unknown function or partial descent, which is X. This is the free variable. So we can say that the fifth PDP is a non-homogeneous PDP because there is a class that does not fit an unknown function or partial descent.

Next, we know that the general form of the linear partial differential equation of the second order is as follows, namely a times dou squared u times dou x squared plus b times dou squared u times dou x dou y plus c times dou squared u times dou y squared minus d equals 0 with a

b and c are the functions of the free variable, namely x and y. While d is the function of x, y, u, ux, or uy. We can write ux as dou u per dou x, so the partial descent of the variable related to u to the free variable x.

and here we can write dou y as dou u per dou y where dou u per dou y is the partial descent of the variable that is not known u to the variable free y

We see that the classification of linear and orderly partial differential equations can be divided into three, namely the first is the hyperbolic partial differential equation. So a partial differential equation is called a hyperbolic partial differential equation.

if the value of b^2 - 4ac is more than 0. What is b? Please note that b is the coefficient of dou^2 u per dou x dou y, while a is the coefficient of dou^2 u per dou x^2 and c is the coefficient of dou^2 u per dou y^2.

If the value of b^2-4ac is more than 0, it is positive, then the PDP is called a hyperbolic PDP.

The second PDP is called a parabolic PDP if the value of b^2 - 4ac = 0. And the third PDP is called an elliptic PDP if the value of b^2 - 4ac is less than 0. So later please pay attention to the values ​​a, b, and c.

which is found in the linear order 2 partial differential equation then calculated the value of b^2-4ac then from the value of b^2-4ac we can determine the type of PDP is PDP like what is it? whether hyperbolic, parabolic, or elliptic so it depends on the value of b^2-4ac

As I have said earlier, if positive, then it will be a hyperbolic PDP, if equal to zero, it is a parabolic PDP, and if the value is less than zero, then it is included in the elliptic PDP. Let's see the example, here is the first example, which is dou squared u per dou t squared equals c squared times dou squared u per dou x squared.

we note that the value of A is negative C squared please note that we bring it to the left side all yes, it means we can write negative C squared dou squared U per dou X squared then added dou squared U per dou T squared this is equal to zero so the value of A is negative C squared the value of B is not there, zero

and the value of c is 1 because here the coefficient of 2 squared u per dot squared is 1 so we can calculate here the value of b squared minus 4ac which is 0 squared minus 4 times minus c squared times 1 which is 0

plus, this is negative times negative, so here it is positive, positive 4c squared, so 0 plus 4c squared, the result is 4c squared. We know here, c squared is a square number, so the result must be positive, multiplied by 4, then the result must also be positive, more than 0.

Therefore, since the value of b^2-4ac is greater than 0, then the PDP is a hyperbolic PDP. Next, for the second example, here dou u/dou t = c^2 dou u^2/dou x^2. We see here, it's the same as before, if this is taken to the left and right, then it will form the differential equation as follows.

So, negative C^2 dou^2 u per dou x^2 plus dou u per dou t equals 0. We note that the value of A is negative C^2, the value of B is 0, and C is also 0 because there is no dou^2 u per dou t^2.

So we can calculate the value of b^2-4ac, which is 0^2-4* ^2-c^2*0, which is 0 + 0. So the result is 0. Because the value of b^2-4ac is equal to 0, then this second PDP is a parabolic PDP.

For the third example, here dou squared u per dou x squared plus dou squared u per dou y squared equals to zero. We see here that the coefficient of dou squared u per dou x squared is one, so the value of A is one.

while here there is no value of dou squared u divided by dou x dou y so here b is 0 and here dou squared u divided by dou y squared is 1 so c is 1 so we can calculate here the value of b squared minus 4ac which is 0 squared minus 4 times a is 1 times c is 1 so here we write 4 times 1 times 1

0 squared, the result is 0. Whereas here, -4 times 1 times 1 is -4. So, 0 minus 4, the result is -4. This is negative, automatically less than 0. Then the PDP at this time is an elliptic PDP. Because the value of b squared minus 4ac is less than 0.

At the end of the course, there is a task that you have to do. Please work on it in the folio paper, then scan and collect it in Google Classroom. That's all for this video. Thank you for your attention. We will meet again at another opportunity. I'm sorry if there is a mistake. Wassalamualaikum warahmatullahi wabarakatuh.