- hace 2 días
Las ecuaciones diferenciales es en, donde intervienen derivadas o diferenciales respecto a una variable independiente.
Ejercicios varios de sencilla resolución, para afianzar conocimientos.
Introducción para el desarrollo de ecuaciones diferenciales en derivadas parciales.
Ejercicios varios de sencilla resolución, para afianzar conocimientos.
Introducción para el desarrollo de ecuaciones diferenciales en derivadas parciales.
Categoría
📚
AprendizajeTranscripción
00:05This video presents a synthesis of higher-order differential equations.
00:10It was previously mentioned that differential equations are equations that relate
00:15a function and its variable or variables with their derivatives or differentials we have type 1
00:20It consists of equations of the form the nth order derivative of y over differential
00:25raised to the nth order is equal to x where x is a function only of x or a constant for
00:31To integrate, we multiply both sides by dx, the derivative raised to the nth order minus 1
00:38is equal to the integrand of the derivative of n of y with respect to x n times all by the differential
00:44of
00:44x is equal to the integrated x times the differential of x plus c sub 1 where it is repeated is n
00:50times less than 1
00:51For this we have the following differential equation, find the solution of the following equation.
00:56The second derivative of x with respect to t is equal to 4 times the sine of 2t. Importantly, we start from the
01:03minor
01:03exponent this is 1 and we integrate the exponent times so we have first derivative is equal
01:08a 4 sine of 2t now we multiply by the differential of t we obtain differential of x is equal to 4 sine
01:15of
01:152t times the differential of t we integrate both sides keeping in mind the form for the first member
01:21that x is equal to the integrand, the first derivative of x times the differential of t is equal to the integral of
01:27x and y
01:27It is also equal to x; similarly, we apply the integration technique when there are constants.
01:32In the integrand, it helps us to clear the constant; solving, we obtain x equals negative 2 times
01:39cosine of 2t plus c sub 1 we have the first integral integrating again at the rate of the exponent applying
01:46The integration technique, when there are constants in the integrand, allows us to
01:51Solving for the constant, we get the answer x equal to negative sine of 2t and c sub 1 plus c sub
01:582
02:00Type 2 differential equations consist of equations of the form second derivative of y
02:05with respect to x is equal to y where y is a function only of x or a constant the method
02:12for
02:12Integration is the following: 1. We write the equation in the differential form of y, which is equal to y.
02:18The differential of x² is multiplied by y and we have y. The differential is equal to
02:25and the differential of x 3 but y quote differential of x is equal to differential of y this transforms
02:32where the equation the variables and and quote are separated 4 integrating we have where the second
02:38member is a function of y 5 taking the square root the variables x and are separated and
02:46We can integrate the following exercise again; it tells us to find the general solution of the following
02:52differential equation the second derivative of y with respect to x is equal to 4 and here we multiply
02:59by an arbitrary variable and quotation mark we obtain we do the integral of the differential over the root
03:04square of y squared plus c over 2 this equals the integrand of 2 we do e over 2
03:10is
03:10equal to c sub 1 integrating this we get natural logarithm of y plus the square root of y squared
03:17plus c sub 1 this is equal to 2x plus c sub 2 making exponents we have as answer e raised
03:262x plus c
03:28sub 4 e raised to the power of negative 2x second order linear differential equations with coefficients
03:35Constants are those of the form second derivative of y plus the constant p times the first derivative plus
03:43the constant whose equals in every linear differential equation with constant coefficients has
03:50The solution is an exponential function, that is, equal to raised to the power of nx, differentiating in the first instance and
03:58From the second derivative, we obtain the ways to establish the equation in squared form.
04:02times raised to the power of nx plus pn raised to the power of nx this is equal to common factor of raised to the power of nx this
04:09multiplies a
04:10This second-degree equation has obtained its roots, which are raised to the power of nx for all x belonging to
04:17These real numbers have distinct roots, so we substitute them into the general equation that
04:23equal to c sub 1 raised to the power of nx plus c sub 2 raised to the power of n sub 2x to obtain a result
04:31with those roots
04:32The following exercise tells us to find the differential equation of the second derivative of x minus the
04:39first derivative minus 2x this equals 0 here identifying the constants p q we make x
04:47equal to e raised to rt differentiating the first derivative we have r e raised to rt performing the second
04:54From the derivative we obtain r squared e raised to rt, we substitute into the differential equation, expanding
05:01And by factoring out the common factor we obtain the roots minus 1 and 2; we substitute these roots into the equation
05:08general and we obtain the answer x equal to c sub 1 raised to the power of -t plus c sub 12 raised to the power of
05:14a 2t for
05:15In case 2, the roots are imaginary, namely one real part and one imaginary part, which is the conjugation of
05:22the roots of the equation n squared plus pn plus q equals 0 analyzing we have raised to n1x
05:31is equal to
05:32raised to the conjugation of a plus b and times x and also its negative part, that is, raised to
05:37n2x is equal to
05:39raised to the conjugation of a minus b and times x n raised to the power of y bx is equal to cosine
05:46of bx minus and sine
05:48Expanding bx, we have raised to x times the cosine of bx, and also raised to x times the sine of bx.
05:56which are
05:57particular solutions and the general solution will be y equal to c 1 raised to x times cosine plus c 12
06:05high
06:05a x times sine of bx now we have the following differential equation the second derivative of s minus
06:132 times the first derivative plus 5s equals 0. For this, we first differentiate the equation x equals
06:21to
06:21raised to rt, its first derivative is r raised to rt, the second derivative we obtain r squared raised to rt
06:30and replacing what was found and substituting into the differential equation and dividing everything by a raised
06:35We have a second-degree equation for rt: r squared minus 2r plus 5, which equals 0
06:42this
06:44equation we apply the formula for solving quadratic equations identifying
06:48Expanding the coefficients, we have 2 plus or minus 4, and taking the imaginary part over 2, we obtain the roots 1 plus 2 and 1.
06:57minus 2 and when we look at these roots we see that they are of the conjugate form where there is a real part a
07:05and another imaginary b, we replace this in the general solution equation and factoring out a common factor
07:12The answer we have is that it is equal to raised to the power of t which multiplies everything makes 1 cosine of 2 t
07:17plus 2 sine of 2 t
07:20If we have a second-degree equation in a differential equations exercise and its roots are
07:26equal this is n 1 n 2 an equality the equation can be written in the following form n square plus
07:33pn
07:33plus one-fourth times p squared this equals a binomial n plus one-half p squared
07:39its roots
07:40will be less than one half p then we have its solutions will be and equal to raised to n 1 x also
07:47and is equal to x that multiplies a raised to the power of n 2 x its general solution will be the replacement of
07:52such
07:52solutions in the general equation and is equal to c 1 raised to n 1 x plus c 12 raised
07:58a n 2 x the next
08:00This exercise tells us to find the general solution to the following second differential equation exercise
08:05derivative of s minus 2 times the first derivative plus s this is equal to zero for this we make s
08:13be the same
08:13a e raised to rt differentiating twice the first derivative we obtain r raised to rt the second derivative
08:20We obtain r squared raised to rt; we substitute these solutions into the differential equation and factoring
08:28By factoring out rt we obtain the second-degree equation r squared minus two
08:35r plus one factoring we get roots equal to one where s is equal to raised to r1 t and s
08:43is equal to t times
08:44raised to r2 t, replacing this in the general equation we find the answer s is equal to c 1 raised
08:51a t plus c 12 raised to t linear differential equations of the form second derivative of ye plus
08:58p times the derivative of ye plus which this is equal to rx in 1 where p and q are constants
09:06that rt x is a
09:07function of the independent variable x or constant steps for the solution 1 let the general solution
09:14and is equal to b the function v we call the complementary function of 1 we denote it by jesus c 2
09:20We determine a particular solution of 1 which we designate as Jesus p 3 the general solution of 1
09:28will be the sum of the complementary function plus the particular solution Jesus g is equal to u plus b in
09:38A linear differential equation will have to be determined to determine the value of y in particular.
09:43an equality when developing the equation according to certain forms the most common are now
09:49We have the exercise of finding the general solution to the differential equation in the following exercise.
09:55second derivative of s plus x this is equal to t plus b for this we make a function is is
10:02equal to e raised
10:03Taking rt twice, we get the first derivative is r raised to the power of rt, the second derivative is r squared
10:10raised to t, once these solutions are obtained, we first make the complementary solution; for this we equate
10:17We set the differential equation to zero and solve it by factoring and taking the square root of negative 1; this gives us
10:23an imaginary root whose value is plus or minus and we observe the conjugate form a plus or minus b and
10:31imaginary once this solution is obtained we have the forms for differential equations with roots
10:37imaginary where x is equal to t times cosine bt also x is equal to t
10:45by sine of bt and
10:48We obtain the complementary function; now for the particular function, the nature of the function will have to be reviewed.
10:54equality of the differential equation so in the differential equation it will no longer be zero in this case it will be equal to
11:00t b
11:01Reviewing the table, we see that the identity that most closely resembles equality is the form a plus bt
11:06in the table it is at x in our case this
11:12This function, by differentiating the first derivative twice, gives us b; the second derivative is zero once these are obtained.
11:20solutions we substitute into the differential equation we solve for the value of b that is b is equal to t plus b
11:27Now that the solutions have been obtained, all that remains is to add the complementary function and the function
11:33In particular, obtaining the answer is equal to one times the cosine of t plus two times the sine of t plus
11:39the
11:40particular function a t plus b nth order differential equations with constant coefficients are a
11:47a specific type of ordinary differential equation that involves derivatives of a function up to a
11:53a certain order n and where the coefficients multiplying these derivatives are constant numbers are expressed as
12:00the following general form a sub n and raised to n plus a sub n minus 1 plus a sub
12:07one more y to sub 0 y this
12:10equal to g function of x where y is the unknown function, first, second, third derivatives
12:20successive from ye to sub n to sub n minus 1 to sub n 0 constant coefficients real numbers or
12:28complex
12:29function of g a known function of x non-homogeneous term main characteristics order the order of the
12:38The equation is determined by the highest order derivative present in this case with constant coefficients.
12:45The numbers multiplying the derivatives do not depend on the independent variable x (linearity).
12:52These equations are linear, which means that the unknown function and its derivatives appear only at
12:57the first power and there are no products between them basic formulas the exact form of the solution
13:04In general, it depends on the roots of the characteristic equation; here we present the cases.
13:10more common real and distinct roots if the characteristic equation has real roots
13:16distinct roots r 1 r 2 r n then the general solution is that function of x is equal to
13:24c 1 e raised
13:25a r 1 x plus c 2 e raised to r 2 x up to nth values where c 1 c
13:312 c n are arbitrary constants
13:34repeated real roots if a root r is repeated m times; complex conjugate roots if there are roots
13:42complex functions of the form alpha plus or minus beta then the corresponding solution involves functions
13:48sinusoidal and cosinusoidal, so we have the following differential equation as an example, find the solution
13:54general of the third derivative of x plus 6 times the second derivative plus 12 times the first derivative plus 8 x
14:02this
14:02equal to 0, so we factor to obtain the roots; in this case we factor by coefficients
14:09separable and we obtain the binomial r plus 2 that multiplies the polynomial r squared plus 4 r plus 4
14:17This gives us two binomials: 1, r plus 2, which multiplies r plus 2, this squared, equating
14:23to 0
14:24We obtain the roots r1 r2 r3 is negative 2 we make x equal to e raised to n
14:31t and x equal to t times e raised
14:34a r sub n by t we designate an auxiliary equation so we have r cubed plus 6 r squared
14:40plus 12 r plus 8
14:42This equals 0. We replace the roots in the general equation, observing that its solution
14:48belongs to equations with equal roots and we obtain the answer by common factoring so x equals e raised
14:56a minus 2 t which multiplies a c 1 plus e 2 by t plus e 3 t squared is
15:02If the video was helpful to you
15:04Don't forget to subscribe
15:05Don't forget to subscribe
Comentarios