Great video. Just an observation, you could easily solve u2 by rewriting the Condition 2 as -u'1*e^(-t) - 2*u'2*e^(-2t) = -[u'1*e^(-t) + u'2*e^(-2t)] - u'2*e^(-2t) . The [ ] equals 0 as we assumed so. Great series!
I've always wondered about VoP: if you don't assume C1, is there a class of solutions seperate from the one you got here, or is C1 a necessary feature of a solution? Thanks for the video. VoP is a mouthful of work!
I noted the same. Just to be clear for the students, the chain rule is when you find the derivative of f(g(t)) that is f'(g(t))*g'(t) and the product rule is when you find the derivative of f(t)*g(t) that is f'(t)*g(t)+f(t)*g'(t) that are different. Nice video, good job.
To be concise, the product rule is a specific case of the chain rule. You can really came up with any "rule" related to the differentiation of a group functions using the chain rule. For example: a "logarithmic" rule for log_u (v) = D(log_u (v))/Du* du/dx + D(log_u (v))/Dv* dv/dx = ln(v)*(-1/u(ln(u))^2)*du/dx + (1/v(ln(u)))*dv/dx = (-v(ln(v))u'+v'(ln(u))u)/(vu(ln(u)^2)) = ln(v^(-vu')*u^(v'u))/(vu(ln(u)^2)) = log_u ( (v^(-vu')*u^(v'u))^(1/vu))/ln(u) = log_u (v^(-u')*u^(v'))/ln(u)
I'm confused. When he takes the derivative of x he uses the product rule, but when he take the derivative of x' he doesn't? EDIT: oh wait, I think I get it now. He assumes the u' terms add to zero, so you don't get any u'' terms in the equation for x''
Dear professor, thank you very much for your lectures. I learned a lot. And I am really interested on the topic, however some of the uploaded videos are unavailable to me (12/45, to be precise). Is it what you intended for them to be or is it a problem with youtube? Thank you again.
I recognize the = 0 step works. Not sure I get when it's OK or not to make the assumption. In particular, did we prive we found all solutions of the equation[?]
I don't know the answer either, but I think the "degrees of freedom" comment provide the direction. My sense is that because we've introduced two unknown functions u_1(t) and u_2(t), we'll end up with two equations that constraint their values i.e. two equations, two unknowns. By being "clever" about the first constraint equation, the system naturally provides the 2nd. If my suspicion is correct, if it were a 3rd order system, we would introduce two constraints and let the system provide the 3rd. The 1st would be as in this video, the second would be Sum( lambda_i * (du/dt)_i * exp(lambda_i * t) ) = 0. And we'd proceed similarly for higher order systems. I hope Prof Steve can point us in a direction to refine this though!
I also have this same question. Why are we able to set this seemingly random condition and not cause a problem? Some insight from the Professor here would be greatly appreciated!
May the Force be with you, GOD bless you Professor Brunton. Thank You !
Great video. Just an observation, you could easily solve u2 by rewriting the Condition 2 as -u'1*e^(-t) - 2*u'2*e^(-2t) = -[u'1*e^(-t) + u'2*e^(-2t)] - u'2*e^(-2t) . The [ ] equals 0 as we assumed so. Great series!
Great lecture...Such a useful scheme to solve 2nd Differential Equations...Thank you very much, Steve...❤🧡💚
This was one of my favorite parts of my differential equations class.
I've always wondered about VoP: if you don't assume C1, is there a class of solutions seperate from the one you got here, or is C1 a necessary feature of a solution? Thanks for the video. VoP is a mouthful of work!
Hi Steve, thanks for all the content, just dont forget to update the differential equation playist.
Hi Steve, only a small correction. Many times in this video you say Chain Rule when what you are using using is the Product Rule.
I noted the same. Just to be clear for the students, the chain rule is when you find the derivative of f(g(t)) that is f'(g(t))*g'(t) and the product rule is when you find the derivative of f(t)*g(t) that is f'(t)*g(t)+f(t)*g'(t) that are different. Nice video, good job.
To be concise, the product rule is a specific case of the chain rule. You can really came up with any "rule" related to the differentiation of a group functions using the chain rule.
For example: a "logarithmic" rule for log_u (v) =
D(log_u (v))/Du* du/dx + D(log_u (v))/Dv* dv/dx =
ln(v)*(-1/u(ln(u))^2)*du/dx + (1/v(ln(u)))*dv/dx =
(-v(ln(v))u'+v'(ln(u))u)/(vu(ln(u)^2)) =
ln(v^(-vu')*u^(v'u))/(vu(ln(u)^2)) =
log_u ( (v^(-vu')*u^(v'u))^(1/vu))/ln(u) =
log_u (v^(-u')*u^(v'))/ln(u)
I'm confused. When he takes the derivative of x he uses the product rule, but when he take the derivative of x' he doesn't?
EDIT: oh wait, I think I get it now. He assumes the u' terms add to zero, so you don't get any u'' terms in the equation for x''
I found what I had missed.
thank You.
Dear professor, thank you very much for your lectures. I learned a lot. And I am really interested on the topic, however some of the uploaded videos are unavailable to me (12/45, to be precise). Is it what you intended for them to be or is it a problem with youtube? Thank you again.
I recognize the = 0 step works.
Not sure I get when it's OK or not to make the assumption.
In particular, did we prive we found all solutions of the equation[?]
I don't know the answer either, but I think the "degrees of freedom" comment provide the direction.
My sense is that because we've introduced two unknown functions u_1(t) and u_2(t), we'll end up with two equations that constraint their values i.e. two equations, two unknowns.
By being "clever" about the first constraint equation, the system naturally provides the 2nd.
If my suspicion is correct, if it were a 3rd order system, we would introduce two constraints and let the system provide the 3rd. The 1st would be as in this video, the second would be Sum( lambda_i * (du/dt)_i * exp(lambda_i * t) ) = 0.
And we'd proceed similarly for higher order systems.
I hope Prof Steve can point us in a direction to refine this though!
I also have this same question. Why are we able to set this seemingly random condition and not cause a problem? Some insight from the Professor here would be greatly appreciated!
Thanks for the good course. Just not sure why suddenly C1 assumption came out of nowhere.
why do you not get u double dot for the second derivative at 7:30?
Did we lose the 2x when rewriting the inhomogeneous eqn in terms of u's?
Beautiful! :)
inhomogenius equations