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.
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 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.
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) }
oh that certainly was an unintended thumbnail
I mean he can say it
That thumbnail called me a slur
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.
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.
I assume the rainbow is in the title in the first seconds of the video, not in the thumbnail @@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.
I definitely need to refresh my mem on Laplace Transforms! Great technique
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) ?
Oh my gosh! First! Hi Dr. Peyam. Love your videos and have used them with my PDE class.
Can you do some videos on controls theory pls?
Why is the thumbnail of this video applying a "Heineken" theme? (those green letters, and that red five-pointed star...)
Nice one but hoped for explicit Lqplace transform of tan
Use wolframalpha
Is that solution valid for all t or only for t > 0 ?
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) }