A little confused in Zero Input response

[zee3b]

Guys I know how to solve the simple ones but this one has complex roots
I have to find the zero input response

(D-1)(D2+1)y(t)=(D2+2)f(t)

y(0)=4 y'(0)=3 y''(0)=3

D2= D square. Y' = first derivative Y''= secondI know the roots are +1, j and -j

But I don't know how to solve or use the j's and their identity

I maybe wrong. If it wasn't for the square inside the parenthesis I would be able to solve it.

Any help would be appreciated. (step by step so i can understand it better) once again I know the whole concept but not the j's

 

[zee3b]

Any one? :(

 

[Mene]

The zero input response refers to the output of a system when there is no input (or forcing function) acting on it. In this case, the system is described by the differential equation (D-1)(D2+1)y(t)=(D2+2)f(t), where D represents the derivative operator.

To find the zero input response, we can use the initial conditions given: y(0)=4, y'(0)=3, and y''(0)=3. These initial conditions represent the state of the system at t=0, before any input is applied.

To solve this differential equation, we first need to find the characteristic equation. This is done by setting the coefficients of the highest derivative term to zero, which in this case is D2+1=0. This gives us the characteristic equation D2+1=0, which has roots of ±j (where j is the imaginary unit, equal to √-1).

Next, we can use the roots of the characteristic equation to find the general solution of the differential equation. This is done by using the formula y(t)=C1e^(r1t)+C2e^(r2t), where r1 and r2 are the roots of the characteristic equation and C1 and C2 are constants to be determined. In this case, the general solution is y(t)=C1e^(jt)+C2e^(-jt).

To find the specific values of C1 and C2, we can use the initial conditions given. Plugging in t=0, we get y(0)=C1+C2=4. Plugging in t=0 for the first derivative, we get y'(0)=jC1-jC2=3. And plugging in t=0 for the second derivative, we get y''(0)=-C1-C2=3. This gives us a system of equations that we can solve to find the values of C1 and C2.

Using these values, we can then find the zero input response by plugging the general solution back into the original differential equation. This gives us the final solution of y(t)=4cos(t)+3sin(t).

In summary, to find the zero input response in this case, we first needed to find the characteristic equation and its roots. Then, we used the initial conditions to find the general solution and determined the specific values of the constants. Finally, we plugged