Aug 11, 2017 · That is, if we fix C₁ as one possible value of C, then since {C₁ ∪ -C₁} ∈ ℝ = C, the sign of the ...
Duration: 4:24
Posted: Aug 11, 2017

Jul 24, 2017 · In the context of the problem we know that C₁ is never going to take imaginary values, but as a ...
Duration: 7:40
Posted: Jul 24, 2017

Mar 16, 2019 · ⌈C₁,₁ = (-1)² * (ei - fh) C₁,₂ = (-1)³ * (di - fg) C₁,₃ = (-1)⁴ * (dh - eg) ⌉ |C₂,₁ = (-1)³ * (bi ...
Duration: 2:48
Posted: Mar 16, 2019

Sep 23, 2014 · ... C₁ and C₂ are only limited by initial conditions (not given in this problem, so you can choose ...
Duration: 7:50
Posted: Sep 23, 2014

Feb 8, 2020 · Ou seja, vai dar o vetor 0,0. Então, basta que um desses números aqui, c₁, c₂ ou c₃ ...
Duration: 17:38
Posted: Feb 8, 2020

Apr 8, 2023 · Com condições iniciais, que eu vou colocar aqui do lado. y(0) = 2 e y'(0) = 3. Para buscar ...
Duration: 6:27
Posted: Apr 8, 2023

Sep 26, 2016 · If we plug x=-2 into the antiderivative pieces, we get 2+C₁ and -2+C₂, for some constants C₁ ...
Duration: 6:53
Posted: Sep 26, 2016

Jun 28, 2018 · In the context of the problem we know that C₁ is never going to take imaginary values, but as a ...
Duration: 7:41
Posted: Jun 28, 2018

May 9, 2023 · Agora vamos fazer a segunda equação, quando a gente substitui "x = 0" na derivada ...
Duration: 9:44
Posted: May 9, 2023

May 27, 2023 · Vamos considerar, também, um vetor "a" com as coordenadas na base B, ou em respeito ...
Duration: 17:55
Posted: May 27, 2023