Ensino Superior ⇒ EDO- Transformada de Laplace Tópico resolvido

[Baguncinha]

y [tex3]^{(4)}[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]^{(4)}[/tex3]

-y=0; y(0)=y''(0)=1, y'(0)=y'''(0)=0

[LPavaNNN]

[tex3]S^4.Y(S)-S^3.y(0)-S^2.y'(0)-S.y''(0)-y'''(0)-Y(S)=0\\Y(s)(S^4-1)=S^3+S\\Y(s)=\dfrac{S(S^2+1)}{(S^2-1)(S^2+1)}=\dfrac{S}{(S-1)(S+1)}\\Y(s)=\dfrac{A_1}{S-1}+\dfrac{A_2}{S+1}\\A_1S+A_1+A_2S-A_2=S\\A_1+A_2=1;A_1-A2=0\\A_1=A_2\\A_1=A_2=\dfrac{1}{2}\\Y(S)=\dfrac{1}{2}.(\dfrac{1}{S-1}+\dfrac{1}{S+1})\\y(t)=\dfrac{1}{2}.(e^t+e^{-t})\\\text{só há exponenciais simples, ou seja, as derivadas pares são iguais à própria função, e as ímpares são iguais entre si}[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]S^4.Y(S)-S^3.y(0)-S^2.y'(0)-S.y''(0)-y'''(0)-Y(S)=0\\Y(s)(S^4-1)=S^3+S\\Y(s)=\dfrac{S(S^2+1)}{(S^2-1)(S^2+1)}=\dfrac{S}{(S-1)(S+1)}\\Y(s)=\dfrac{A_1}{S-1}+\dfrac{A_2}{S+1}\\A_1S+A_1+A_2S-A_2=S\\A_1+A_2=1;A_1-A2=0\\A_1=A_2\\A_1=A_2=\dfrac{1}{2}\\Y(S)=\dfrac{1}{2}.(\dfrac{1}{S-1}+\dfrac{1}{S+1})\\y(t)=\dfrac{1}{2}.(e^t+e^{-t})\\\text{só há exponenciais simples, ou seja, as derivadas pares são iguais à própria função, e as ímpares são iguais entre si}[/tex3]