Great explenation. Much more compact than what i heard at school or university. They usually start with a special case of single sine or cosine expression. Build up to A sin(wt)+ B cos(wt) with a last minute turn to e functions over Eulers formula. Where they usually lose everybody in the hurry. This general picture/strategy tailored down to the special case works way better for me now. ^^ but i hold an Eng. degree now so this stuff is not new for me anymore.
Since sin and cos are basically just special cases of the exponential function, it's definitely simpler to start there. Exponentials are easier to conceptualize when using derivatives anyway.
Great lecture! I'm particularly interested to know more about your technical setup. Is this a transparent whiteboard? But then, the writing is not reflected like in a mirror! ^^
So because differentiation is a linear operator, the homogenous solution is like the nullspace of that operator. Just like with linear transformations, the full set of solutions is given by the particular solutions plus the nullspace, because the transformation collapses the nullspace to nothing.
Very thanks ❤❤❤...As I learned from the problem in 19:01, the values of c1, c2, A, and B depend on the value of \omega. is that right? It is my solution for A and B, when I put x(t) = Acos(wt)+Bsin(wt) in the ODE: [A;B] = [2-w^2 3w; -3w 2-w^2]^-1 * [1;0]
Great explenation. Much more compact than what i heard at school or university. They usually start with a special case of single sine or cosine expression. Build up to A sin(wt)+ B cos(wt) with a last minute turn to e functions over Eulers formula. Where they usually lose everybody in the hurry.
This general picture/strategy tailored down to the special case works way better for me now. ^^ but i hold an Eng. degree now so this stuff is not new for me anymore.
Since sin and cos are basically just special cases of the exponential function, it's definitely simpler to start there. Exponentials are easier to conceptualize when using derivatives anyway.
Marvelous! It's a pleasure to watch your videos.
That was flawless, sir. Thanks a lot!
Great lecture! I'm particularly interested to know more about your technical setup. Is this a transparent whiteboard? But then, the writing is not reflected like in a mirror! ^^
Write on glass and then flip the recorded video horizontally to get the text back in the right direction
@@AlistairLynn That makes sense if he is only writing on glass. However, in some other lectures, he also projects python code on the glass.
So because differentiation is a linear operator, the homogenous solution is like the nullspace of that operator. Just like with linear transformations, the full set of solutions is given by the particular solutions plus the nullspace, because the transformation collapses the nullspace to nothing.
Very thanks ❤❤❤...As I learned from the problem in 19:01, the values of c1, c2, A, and B depend on the value of \omega. is that right?
It is my solution for A and B, when I put x(t) = Acos(wt)+Bsin(wt) in the ODE:
[A;B] = [2-w^2 3w; -3w 2-w^2]^-1 * [1;0]
Thank you 💗