ANÁLISE ATPS DE SINAIS E SISTEMAS

ATPS ANÁLISE DE SINAIS E SISTEMAS

y” = 0

y”+16y = 0

y”+6y’+9y = 0

y”+16y’ = 0

y”+16y = 0 ; y(0) = 1 y’(0) = -4

y”+4y’+4y = 0 ; y(0) = 1 ; y’ = 0

y”+5y’+6y = 0 ; y(0)= 1 ; y’(0) = -1

y”+y’ = 0 ; y(0) = 2 ; y’(0) = -1

1) - y” = 0

a=1 x=(-0±√(0^2-4.1.o))/(2.1) y = C1e^0 + C2xe^0

b=0 y = C1 + C2x

c=0 x = 0

2) - y”+16y = 0

a=1 y=(-b±√(0^2-4.1.16))/2.1 α =0

b=0 y=(-b±√64)/2a β=j4

c=16 y=(-0±j8)/2

y = C1.e^αx.cos⁡〖(βx)+C2e^(αx ) 〗 〖.sin〗⁡〖(βx)〗

y = C1.e^0.cos⁡〖(4x)+C2e^(0 ) 〗 〖.sin〗⁡〖(4x)〗

y = C1.cos⁡〖(4x)+C2〗 〖.sin〗⁡〖(4x)〗

3) y”+6y’+9y = 0

a = 0 y =(-6±√(6^2-4.1.9))/2.1

b = 6 y =(-6±√0)/2.1

c = 9 y =-3

y=C1.e^r1x+C2.xe^r2x

y=C1.e^(-3x)+C2.xe^(-3x)

4) y’’+16y’ = 0

a = 1 y=(-b±√(〖16〗^2-4.1.0))/2.1 y=C1.e^0x+C2.e^(-16x)

b = 16 y=(-16±√(〖16〗^2 ))/2.1 y=C1.+C2.e^(-16x)

c = 0 y=(-16±16)/2.1

y1= 0 y2=-16

5) y’’+16y = 0 Condições : y(0) = 1 ; y’(0) = -4

a = 1 y=(-b±√(0^2-4.1.16))/2.1 α =0

b = 0 y=(-b±√64)/2a β=j4

c = 16 y = C1.e^αx.cos⁡〖(βx)+C2e^(αx ) 〗 〖.sin〗⁡〖(βx)〗

y = C1.e^0.cos⁡〖(4x)+C2e^(0 ) 〗 〖.sin〗⁡〖(4x)〗

y = C1.cos⁡〖(4x)+C2〗 〖.sin〗⁡〖(4x)〗

para y=(0)=1

C 1 cos⁡〖(4.0)+C2 sin⁡〖(4.0)〗 〗 =1 C1=1

y=1.cos⁡〖(4x)+C2 sin⁡〖(4x)〗 〗

y^'=-4 sin⁡〖(4x)+C2.4 cos⁡〖(4x)〗 〗

Para y=(0)=-4

-4 sin⁡〖(4x)+C2.4 cos⁡〖(4x)〗 〗 C2 = -1

-4 sin⁡〖(4.0)+C2.4 cos⁡〖(4.0)〗 〗=-4

y=1.cos⁡〖(4x)-1 sin⁡〖(4x)〗 〗

y=cos⁡〖(4x)-sin⁡〖(4x)〗 〗

6) y”+4y’+4y = 0 ; y(0) = 1 ; y’ = 0 Utilizando regra da cadeia

a = 1 □∆=(-4±√(4^2-4.1.4))/2.1 e^(-2x) = 〖(e〗^(u^' )).(v^') u= -2x

b = 4 ∆ = -2 〖 e〗^(u ).-2

c = 4 -2e^u

-2e^(-2x)

y=C1e^(-2.0)+C2xe^(-2.0) y^'=C1.(-2.e^(-2x))+C2(-2e^(-2x))

y=C1.1