00:00 - special class of 1st order ODE w/ explicit solutions 00:17 - definition of first order linear equations 01:19 - y appearing just by itself, else is the function of t. 01:40 - counter-example, non-linear 02:19 - example with LHS and RHS issues by changing signs of p(t) 03:30 - solving first order linear ODE’s 04:48 - what if DE is multiplied with μ(t) ? 06:45 - condition for DE, “to really have equality” with (μ(t) * y)’ 07:46 - condition turns into a separable equation 09:25 - μ is found, 09:40 - comment: redundancy of c, simple form, key analogy, non-zero c 10:30 - μ is integrating factor 11:12 - example, all solutions 12:40 - whether to put +c 14:59 - essentiality of c after integration, to find all solutions to DE 15:44 - example w/ a given initial condition, and solution interval is sought 17:20 - conversion into general form 19:25 - concerns about |t|^2, (intervals which t is non-zero is to be separately treated) 22:30 - initial value checks 23:39 - largest interval of definition, relevant interval is determined w/ initial value check for t 25:05 - example, which needs a simple change to make it linear 26:20 - swapping dependent and independent variable 28:40 - no absolute value since y=0 is problem point 30:15 - integrate with respect to y 30:50 - solving quadratic eq. In order to write y in terms of x 31:28 - checking whether y=0 is a solution, keeping track of all division steps 32:34 - exponential growth and decay, constant p 35:37 - example w/ constant q, and an interpretation 40:00 - a general definition of growing exponential and decaying exponential
Hocam, at 32.00 how can we say we divided it by y so we need to check y = 0 when we flipped the equation. I couldn't see the division by y. And at 25.45 can't we say (-inf, 0) U (0,+inf)
When we flip the equation algebraically this means we are looking at 1 over both sides. But then we will be dividing 1 by y which is in the numerator of the original right hand side. This implicitly assumes that y is not 0, so it should be checked.
The interval of definition should be a connected interval. The other disconnected piece is not meaningful from the standpoint of differential equations, so we ignore it.
For differential equations, a connected interval containing the initial point is relevant. Another, disconnected interval may seem ok for the function, but it is physically irrelevant for the differential equation.
The term "largest" is used in the sense of set inclusion; if one set is a subset of another, then the set containing the other one is considered "larger". It is not necessarily measured by length, etc. In this question, one of these two infinite intervals does not contain the initial point, so we have to discard it.
Hocam, at 23.30 I tested y(1) = 1/2 after leaving y alone and making C absorb (I don't know the term for it) 1/2 and t^2. Is this an okay way of solving it? If not why?
Hocam bir şey sorucam dakika 20.07de çözdüğün soruda interprating factor fonksiyonunda e^2lnt buluyoruz buarada mutlak olmadığı için; dakika 23.51de çözdüğünüz largest interval da direkt olarak t > 0 dememiz gerekmez mi? fonksiyonun hangi kısımda olduğuna bakmaksınız.
3 года назад+1
Integrating factoru e^2ln|t| şeklinde yazsaydık da olurdu, netice değişmiyor. 19:22 de bunu açıklıyor aslında. Yani initial condition y(-1)=1/4 de verilebilirdi mesela, o durumda da bu integrating factor çalışırdı ve (-infty,0) aralığında tanımlı bir çözüm fonksiyonu bulurduk.
At 32.00 Even if we swap the dependent and independent variables, why do we find out y as a result. Despite of swapping variables we didnt make any change.
When we flip the equation algebraically this means we are looking at 1 over both sides. But then we will be dividing 1 by y which is in the numerator of the original right hand side. This implicitly assumes that y is not 0, so it should be checked.
When we flip the equation algebraically this means we are looking at 1 over both sides. But then we will be dividing 1 by y which is in the numerator of the original right hand side. This implicitly assumes that y is not 0, so it should be checked. To actually check it, substitute y=0 to the DE and see if the two sides agree.
@@aliulasozgurkisisel4991 hocam, can you explain it a little bit widely.I did not get it as well. How we actually write that y=0 is also a solution. The rest is satisfactory, only this part remained unclear.
@@narimannovruzov396 y=0 should we checked, since at some point we divided by y. We can check it by plugging y=0 in the equation directly and see if the left and right hand sides match.
Dear Oussama, the justification is the theorem that the derivative of an inverse function is the multiplicative inverse of the derivative of the original function. The notation is designed so as to capture this fact.
The antiderivative of 1/t is ln|t|. So it gives different answers for t>0 and t>0. The equation itself has a bad point at t=0: at t=0 p(t) is not continuous. For this reason, the two intervals (0,+infty) and (-infty,0) require separate treatment.
00:00 - special class of 1st order ODE w/ explicit solutions
00:17 - definition of first order linear equations
01:19 - y appearing just by itself, else is the function of t.
01:40 - counter-example, non-linear
02:19 - example with LHS and RHS issues by changing signs of p(t)
03:30 - solving first order linear ODE’s
04:48 - what if DE is multiplied with μ(t) ?
06:45 - condition for DE, “to really have equality” with (μ(t) * y)’
07:46 - condition turns into a separable equation
09:25 - μ is found,
09:40 - comment: redundancy of c, simple form, key analogy, non-zero c
10:30 - μ is integrating factor
11:12 - example, all solutions
12:40 - whether to put +c
14:59 - essentiality of c after integration, to find all solutions to DE
15:44 - example w/ a given initial condition, and solution interval is sought
17:20 - conversion into general form
19:25 - concerns about |t|^2, (intervals which t is non-zero is to be separately treated)
22:30 - initial value checks
23:39 - largest interval of definition, relevant interval is determined w/ initial value check for t
25:05 - example, which needs a simple change to make it linear
26:20 - swapping dependent and independent variable
28:40 - no absolute value since y=0 is problem point
30:15 - integrate with respect to y
30:50 - solving quadratic eq. In order to write y in terms of x
31:28 - checking whether y=0 is a solution, keeping track of all division steps
32:34 - exponential growth and decay, constant p
35:37 - example w/ constant q, and an interpretation
40:00 - a general definition of growing exponential and decaying exponential
Hocam, at 32.00 how can we say we divided it by y so we need to check y = 0 when we flipped the equation. I couldn't see the division by y.
And at 25.45 can't we say (-inf, 0) U (0,+inf)
When we flip the equation algebraically this means we are looking at 1 over both sides. But then we will be dividing 1 by y which is in the numerator of the original right hand side. This implicitly assumes that y is not 0, so it should be checked.
The interval of definition should be a connected interval. The other disconnected piece is not meaningful from the standpoint of differential equations, so we ignore it.
Great explanation, thank you
hocam, I did not understand that at 24.29 why do we choose one of the intervals?
For differential equations, a connected interval containing the initial point is relevant. Another, disconnected interval may seem ok for the function, but it is physically irrelevant for the differential equation.
25:03 de t=0 da interval'in içinde değil mi? belki Y(t)=1/2 sabit fonksiyon olamaz mı?
At 24:40, I didn't understand how did we choose the largest open interval. Can we compare size of these two intervals ,(0,+infty) and (-infty,0)?
The term "largest" is used in the sense of set inclusion; if one set is a subset of another, then the set containing the other one is considered "larger". It is not necessarily measured by length, etc. In this question, one of these two infinite intervals does not contain the initial point, so we have to discard it.
Hocam, at 23.30 I tested y(1) = 1/2 after leaving y alone and making C absorb (I don't know the term for it) 1/2 and t^2. Is this an okay way of solving it? If not why?
c cannot absorb t^2. The term t^2 is not a constant
Hocam bir şey sorucam dakika 20.07de çözdüğün soruda interprating factor fonksiyonunda e^2lnt buluyoruz buarada mutlak olmadığı için; dakika 23.51de çözdüğünüz largest interval da direkt olarak t > 0 dememiz gerekmez mi? fonksiyonun hangi kısımda olduğuna bakmaksınız.
Integrating factoru e^2ln|t| şeklinde yazsaydık da olurdu, netice değişmiyor. 19:22 de bunu açıklıyor aslında.
Yani initial condition y(-1)=1/4 de verilebilirdi mesela, o durumda da bu integrating factor çalışırdı ve (-infty,0) aralığında tanımlı bir çözüm fonksiyonu bulurduk.
At 32.00 Even if we swap the dependent and independent variables, why do we find out y as a result. Despite of swapping variables we didnt make any change.
When we flip the equation algebraically this means we are looking at 1 over both sides. But then we will be dividing 1 by y which is in the numerator of the original right hand side. This implicitly assumes that y is not 0, so it should be checked.
I could not understand that at 28.26 why we did not write ln|y| instead of lny
Essentially, it is because both would give us integrating factors, but finding one of them Is enough.
Hocam , in 32:00 how did we check y=0 was also a solution?
When we flip the equation algebraically this means we are looking at 1 over both sides. But then we will be dividing 1 by y which is in the numerator of the original right hand side. This implicitly assumes that y is not 0, so it should be checked. To actually check it, substitute y=0 to the DE and see if the two sides agree.
@@aliulasozgurkisisel4991 hocam, can you explain it a little bit widely.I did not get it as well. How we actually write that y=0 is also a solution. The rest is satisfactory, only this part remained unclear.
@@narimannovruzov396 y=0 should we checked, since at some point we divided by y. We can check it by plugging y=0 in the equation directly and see if the left and right hand sides match.
Hocam what is the justification for inverting the fraction at 26:30? Isn't dy/dx supposed to be just an operator and not an actual fraction?
Dear Oussama, the justification is the theorem that the derivative of an inverse function is the multiplicative inverse of the derivative of the original function. The notation is designed so as to capture this fact.
@@aliulasozgurkisisel4991 Hocam, how do we know the existence of the inverse function of y?
At 20.29 why do we need to thread t0 seperately?
The antiderivative of 1/t is ln|t|. So it gives different answers for t>0 and t>0. The equation itself has a bad point at t=0: at t=0 p(t) is not continuous. For this reason, the two intervals (0,+infty) and (-infty,0) require separate treatment.
Hocam, videolarda anlattığıniz yerlerin (yani kendi yazdığınız yerler) notunu da pdf şeklinde atabilirmisiniz ?
OK, METUClass'a koymaya çalışacağım.