Convolution and ODE

[@ariel_haymarket]

oh that certainly was an unintended thumbnail

[@firelow]

I mean he can say it

[@eduardolezcano4323]

DR Peyam, eres genial mi amigo!

[@drpeyam]

Gracias 😊

[@omgopet]

Dude, that star in the thumbnail that looks a lot like "A" and the rainbow flag around the title make this video the greatest accidental shitpost of the year so far.

[@yurenchu]

What "rainbow flag"? All I'm seeing in the thumbnail is a red five-pointed star combined with two green letters; and that reminds me only (and very strongly) of Heineken beer.

[@firelow]

I assume the rainbow is in the title in the first seconds of the video, not in the thumbnail ​@@yurenchu

[@yurenchu]

​​@@firelow Ah yes, I see it now. Thanks. It reminds me more of prismatic color shifting, than any flag (especially since the colors vary along the horizontal axis, instead of the vertical axis); but even if the maker was inspired by the rainbow flag, I don't have a problem with that.

[@koii55]

That thumbnail called me a slur

[@garyhuntress6871]

I definitely need to refresh my mem on Laplace Transforms! Great technique

[@hugohugo37]

Oh my gosh! First! Hi Dr. Peyam. Love your videos and have used them with my PDE class.

[@yurenchu]

The challenging follow-up question: what are the values of y(t) , y'(t) and y"(t) when t = π/2 + πk (for any integer k) ?

[@thebeerwaisnetwork8024]

Can you do some videos on controls theory pls?

[@md2perpe]

Is that solution valid for all t or only for t > 0 ?

[@yurenchu]

I think it's valid for all (real) t. It's quite easy to verify that the solution is valid for t=0 (because with the given boundary conditions, it fits the original ODE). For negative t , substitute t = -x (for positive x) and τ = -θ into the video's solution and derive that for negative t , the formula is equivalent to y(t) = (e^[2t]) * { ∫ (t+θ)*(e^[2θ])*tan(θ) dθ , from θ=0 to θ=(-t) } Then determine y'(t) and y"(t) , and insert them into the ODE to verify that this solution fits. Hint: I used y(t) = (e^[2t]) * (F(t) + G(t)) where F(t) = t * { ∫ (e^[2θ])*tan(θ) dθ , from θ=0 to θ=(-t) } G(t) = { ∫ θ*(e^[2θ])*tan(θ) dθ , from θ=0 to θ=(-t) }

[@yurenchu]

Why is the thumbnail of this video applying a "Heineken" theme? (those green letters, and that red five-pointed star...)

[@BrendanLawlor-m5n]

Nice one but hoped for explicit Lqplace transform of tan

[@lih3391]

Use wolframalpha