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Las ecuaciones diferenciales es en, donde intervienen derivadas o diferenciales respecto a una variable independiente.
Ejercicios varios de sencilla resolución, para afianzar conocimientos,
Ejercicios varios de sencilla resolución, para afianzar conocimientos,
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00:06This video demonstrates how to formulate differential equations, in this case, first-order differential equations.
00:11At this level, differential equations are those where derivatives or differentials with respect to
00:16an independent variable, and can be written as follows. The differential function m
00:22x, plus n functions of the differential y equal to zero. Where m, n, are functions of x, y, of
00:29the equations
00:30differentials that belong to we have the types to know.
00:36Type 1, separable equations. These are differential equations whose terms are
00:42They can be arranged in the form: Function of x times differential of x, plus function of y times differential
00:49where y is a function of x only, and y is a function of y only. This procedure
00:55This is called separation of variables, and its solution is obtained by direct integration. The integrand
01:00of the function of x, plus the integrand of the function of y equal to c, where c is a constant
01:05arbitrary. We have the following resolution rules. Step 1, remove denominators, if
01:14The equation contains derivatives; all terms are multiplied by the differential of the
01:19independent variable. Step 2, the differentials are factored out if then
01:25The equation takes the form: XY times the differential of x, plus, XY times the differential of
01:32and equal to zero. Where x, x', are functions of x only and y, y', are functions
01:38of y can only be reduced to the form. Function of x times differential of x, plus function
01:43of y by the differential of y equal to zero, dividing everything by x quote y. Step 3, each is integrated
01:50separate parts according to the form, the integrand of the function of x, plus the integrand of
01:55The function y equals c, where c is an arbitrary constant. Thus we have the following exercise,
02:02Find the general solution of the following differential equation, XY by the differential
02:07of x, minus, in parentheses, 1, plus x, squared times the differential of y, equals
02:13to zero. Moving the binomial to the second member, solving for variables, then
02:18We take the differential of a function b, equal to, in parentheses, 1, plus x, squared,
02:25By solving for the integral, we obtain, by substitution, the integral, which is one-half. Integrating from
02:31the differential of b, over b, is equal to the integrand of the differential of y over y.
02:38Moving everything to the first member of the equation, and equating it to c, according to the form of integration,
02:44Integrating we have, one half times the natural logarithm of the binomial, 1, plus x squared,
02:49The negative natural logarithm of y equals c; here it is important to keep in mind the integral
02:54of the differential of b over the same function b, is equal to the natural logarithm of b, plus
02:59c, and this is equal to a product of b by the constant c, according to the rules of natural logarithm.
03:10Multiplying the second member by 2, the natural logarithm of y, plus e, taking into account that
03:15The natural logarithm of e is equal to 1, so we have the following answer: the binomial of 1 plus
03:21x squared is equal to y squared times c.
03:24Homogeneous equations, it is said that the differential equation, the differential function m x, plus
03:30n, a function of the differential and equal to zero, is homogeneous when m, n, are homogeneous functions
03:35of x, y of the same degree, that is to say that they verify the following identity.
03:40The function of x, y multiplied by lambda is equal to lambda raised to the power of n values, times the
03:45function of x, y.
03:47The rule is as follows and is solved by making the substitution y equal to u by x.
03:51This will give a differential equation in u, x where the variables are separable and
03:56proceed to solve according to the rules of separable equations
03:59studied.
04:00We have the example of solving the following differential equation, binomial of 2x, plus, and by the
04:06differential of x, plus binomial of 2x, plus 3y, times the differential of y equal to zero.
04:13First we multiply by lambda, recognizing the functions of m, n, also doing and
04:18equal to u times x, this we perform the differential of the product.
04:23Substituting into the differential equation y and the differential in y, we have, multiplying
04:28and reducing we have x multiplying the polynomial 1, plus, 4u, minus 3u squared,
04:34by the differential of x, plus x, squared, which multiplies the binomial 2, minus
04:403u times the differential du, this equals zero.
04:44We divide all this by the polynomial, x squared, which multiplies 1, plus, 4u, minus
04:503u squared, in order to have separable functions.
04:54Integrating this and taking into account the way to do the integration by integrating the
04:58function of x times the differential of x, plus, the integrand of the function y, times the differential
05:04and this equals c.
05:05The second integral is solved by substitution, we make a function z equal to 1, plus, 4u,
05:11negative 3u squared, performing the differential and solving we get one half
05:16by the differential of z, is equal to 2, minus 3u, this by the differential of u, solving
05:22the integrals and simplifying by natural logarithm.
05:25We have the natural logarithm of x squared, z equals c.
05:30Taking into account that the natural logarithm of e is equal to 1, we obtain x squared
05:35z, is equal to c, here we replace the function z, and also y, and simplifying we get
05:41The answer is x squared, plus 4xy, minus 3y squared, this equals c.
05:48Linear equations of type 3, a first-order equation is an equation that is linear
05:52both in the dependent variable and in its derivative and has the form, derivative of and
05:56with respect to x, plus, and by the function px, is equal to qx at 1.
06:02Where p, q are functions of y only or constants as well.
06:07The derivative of x with respect to y, plus x times the function fx, is equal to jy.
06:14From equation 1 to its standard form, multiplying everything by the differential of x yields
06:19obtains, differential of y, plus y by the function of px, by the differential of x is
06:25equal to qx, the differential of y. We have the integrating factor, the integrating factor of an equation
06:32differential, is defined as a function that, when multiplied by a differential equation
06:37If it's not exact, you can convert it into an exact differential equation. Its antiderivative and equality
06:42They also appear in other exercises.
06:46Rule for solving linear equations.
06:48Step 1. Put equation 1 into standard form.
06:53Step 2. Obtain the integrating factor.
06:57Step 3. Apply the integrating factor to the equation in its standard form.
07:02Step 4. Solve the resulting equation.
07:05For this we have the following exercise: find the solution to the differential equation
07:10x times the derivative of y with respect to x, minus 2y equals 2x.
07:14Here we observe the form where p, q are functions and constants without derivatives or differentials,
07:20First, we multiply everything by the differential of x, and divide by x in order to take the
07:25In the form already studied, and we have the function p of x is equal to negative 2 over x, we obtain
07:31The differential of y, minus 2y over x times the differential, is equal to 2 times the differential of x.
07:38Now we develop the integrating factor, which serves to multiply an equation
07:43A non-exact differential equation allows you to convert it into an exact one in its variables, to make it easier
07:47its solution. That is, raised to the integral of the function p in x times the differential, solving
07:54and developing the natural logarithm, keeping in mind that the natural logarithm of e is equal to
07:58a 1, we have as an answer 1 over x squared.
08:06Multiplying 1 over x squared by the differential of y, minus 2y over x by
08:12The differential is equal to 2 times the differential of x.
08:16We have the differential of y over x squared, minus 2y times the differential of x, over x raised
08:22cubed, this equals 2 times the differential of x over x squared.
08:27Now we apply the product of two functions, this to find out how the
08:32product of the differential of the two terms, in the differential of y over x squared,
08:36negative 2y times the differential of x, over x cubed, this equals 2 times the differential
08:42of x over x squared.
08:45We verify that it is x raised to the power of negative 2 times y, we apply integrals to both
08:50members, in the first member is the integrand of the differential of x raised to the negative 2
08:55by y, it gives us the same function.
08:59Integrating we get y over x squared, plus 2 over x this equals c, solving we get
09:04The answer is y equals c times x squared, minus, 2x.
09:08Type 4, equations that can be reduced to linear form.
09:13If a variable appears in the second member of a linear differential equation with a high power
09:18a n values or nth values, is also called the Bernoulli equation.
09:23Where if n equals 1 it is solved by separable variables, if n is not equal to 1, it can be reduced
09:29To form 3 we do by substitution, z function is equal to y raised to negative n, plus 1, this
09:35implies
09:35The differential of z is equal to the binomial that multiplies y raised to the negative n, by the
09:40differential of y.
09:42We have the exercise to find the differential equation of the derivative of y with respect to x, plus, y over
09:47x, is equal to y cubed.
09:51First, we convert the equation to its standard form, then we do it by substitution.
09:55Thus, for the function z equals y raised to the power of -2, taking the differential z, we get
10:00It has, and the integrating factor is negative 2 over x.
10:10We take the integrating factor, we get 1 over x squared.
10:15Now multiplying 1 over x squared, we get the differential of z over x squared, minus
10:212z over x cube, times the differential x, is equal to negative 2 differential of x over x
10:26square.
10:29Integrating we get, but we have z function equal to 2x, plus, c times x squared,
10:35We also have z equal to y raised to the power of -2.
10:39Expanding and substituting, we obtain the answer cx squared and squared,
10:43plus, 2xy squared, minus 1 equals 0.
10:48That's all, if you found the video helpful, don't forget to subscribe.
10:52Thank you!
10:52Thank you!
10:53Thank you!
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