This is the best explanation I’ve ever seen for this type of diff eq! Very deep and thorough explanation! I really enjoy your videos. Please keep them coming!
Without watching (yet) I will just say: « because exp ([plus or minus]i2y) has a second derivative which is the additive inverse of the function. (Thé factor of four means that a factor of two must be in the argument of the exponential.) ». I like your solution
@@manfredwitzany2233 That is something i already knew, in math you more would like to have a generell solution independend of reality laws to describe abstract objects. But like accepted if you suppose the complex part to be 0 then you get the solution from the Video and it would even describe reality close related problems. But thank's anyway :) (Sorry if i misspelled something, i am not a native speaker. ^^)
@@99selfmade21 Your argument is correct. But in maths we often look for real solutions using complex numbers. Think about complex analysis for solving integrals of rational functions. This is much easier than solving by fractions.
Intersting way to solve. I had to solve many of such differential equations (so called harmonic oscillator). The way I used is a bit more easy. As the differential equation is linear, the general solution can be written as real part of an exponential function. y=Re(A exp(Bt+C)) with unknown complex constants A, B, C. Due to the +C in the exponent A and C can be taken real. This function can be differentiated twice to get y"=AB² exp(Bt+C). These functions for y and y" can be substituted in the differential euqation and you will get: Re((AB² + 4A) exp(Bt+C)) = 0. This equation must be valid for each value of t, which only can be fulfilled if the factor of the exp function equals zero. This leads to the equation B²=-4 or B=+-2i. A and C are arbirtary real values, which only can be determined by means of starting values of y and y'. Using the exp-cos-sin-identity and the fact that only the real part has to taken, the result is: y=A cos(2t+C) being eqivalent to your solution. A is called amplitude and C is called the phase of the oscillation.
@@drpeyam To fulfill the differential equation, y and y" must be proportional with each other. Otherwise the sum would not be zero for every t. There are only two functions fulfilling this, namely the exponetial function and the zero function. The later one is just the spacial case A=0. Therefore the exponetial function is the most general solution. The statement holds even there is some linear y' term. Nevertheless there is a problem with this argument, as it does not seem beeing evident for a sum of three terms.
Really very nice work good representation and nice discussion, but I still one of the vans of the differential operator method which is more easier in applications for homogeneous HODE and also the extension for the inverse operator in dealing with the nonhomogeneous HODE
Hey Dr Peyam I love your videos and enyojed the video "Differential Equations the cool way" but could you please explain the method of the differential opperator and explain why it works and why you're allowed to factor stuff. Thanks for your vids:)
You showed that if the imaginary part of the solution is 0 then B2=-A2 and B1=A1. But don't you need to prove the opposite direction too? How do you know that B2=-A2 and B1=A1 implies that the imaginary part of the solution is 0? Edit: i looked at it again and the opposite direction is trivial, just thought it would be worth mentioning out unless i'm missing something.
This is the best explanation I’ve ever seen for this type of diff eq! Very deep and thorough explanation! I really enjoy your videos. Please keep them coming!
People like you inspire me to take up Physics/Math as a career option. Regards from a high school sophomore
Without watching (yet) I will just say: « because exp ([plus or minus]i2y) has a second derivative which is the additive inverse of the function. (Thé factor of four means that a factor of two must be in the argument of the exponential.) ». I like your solution
Do we just suppose, that y has to be a real function? Or is there a reason why?
We are looking for real solutions
@@drpeyam aahh thanks. ^^ I bet then, I misheard it in the video thank you.
This is the differential equation for a harmonic ocsillator. y is the postion of the oscillating particle. A complex y would be strange.
@@manfredwitzany2233 That is something i already knew, in math you more would like to have a generell solution independend of reality laws to describe abstract objects. But like accepted if you suppose the complex part to be 0 then you get the solution from the Video and it would even describe reality close related problems. But thank's anyway :) (Sorry if i misspelled something, i am not a native speaker. ^^)
@@99selfmade21 Your argument is correct. But in maths we often look for real solutions using complex numbers. Think about complex analysis for solving integrals of rational functions. This is much easier than solving by fractions.
Intersting way to solve. I had to solve many of such differential equations (so called harmonic oscillator). The way I used is a bit more easy. As the differential equation is linear, the general solution can be written as real part of an exponential function. y=Re(A exp(Bt+C)) with unknown complex constants A, B, C. Due to the +C in the exponent A and C can be taken real. This function can be differentiated twice to get y"=AB² exp(Bt+C). These functions for y and y" can be substituted in the differential euqation and you will get: Re((AB² + 4A) exp(Bt+C)) = 0. This equation must be valid for each value of t, which only can be fulfilled if the factor of the exp function equals zero. This leads to the equation B²=-4 or B=+-2i. A and C are arbirtary real values, which only can be determined by means of starting values of y and y'. Using the exp-cos-sin-identity and the fact that only the real part has to taken, the result is: y=A cos(2t+C) being eqivalent to your solution. A is called amplitude and C is called the phase of the oscillation.
How do you know your first statement is true?
@@drpeyam To fulfill the differential equation, y and y" must be proportional with each other. Otherwise the sum would not be zero for every t. There are only two functions fulfilling this, namely the exponetial function and the zero function. The later one is just the spacial case A=0. Therefore the exponetial function is the most general solution. The statement holds even there is some linear y' term. Nevertheless there is a problem with this argument, as it does not seem beeing evident for a sum of three terms.
Really very nice work good representation and nice discussion, but I still one of the vans of the differential operator method which is more easier in applications for homogeneous HODE and also the extension for the inverse operator in dealing with the nonhomogeneous HODE
Hey Dr Peyam I love your videos and enyojed the video "Differential Equations the cool way" but could you please explain the method of the differential opperator and explain why it works and why you're allowed to factor stuff. Thanks for your vids:)
Love your videos.
Which university do you teach in doc?
UC Irvine
You are amazing
Wow ? I am fond of diff.equation السلام عليكم
You showed that if the imaginary part of the solution is 0 then B2=-A2 and B1=A1. But don't you need to prove the opposite direction too? How do you know that B2=-A2 and B1=A1 implies that the imaginary part of the solution is 0?
Edit: i looked at it again and the opposite direction is trivial, just thought it would be worth mentioning out unless i'm missing something.
I had a stroke reading the thumbnail
why do we need to put an I
We need operators. y is not an operator, but Iy is
Didn't know mark ruffalo did math
Hahaha
Lol
Cool