The solution to the ordinary differential equation y˙(t)+2y(t)=2,y(0)=0,𝑦˙(𝑡)+2𝑦(𝑡)=2,𝑦(0)=0, isy(t)=t𝑦(𝑡)=𝑡y(t)=1−e−2t𝑦(𝑡)=1−𝑒−2𝑡y(t)=2−2e−2t𝑦(𝑡)=2−2𝑒−2𝑡y(t)=1−2e−2t

The question seems to be asking for the solution to the ordinary differential equation y'(t) + 2y(t) = 2, with the initial condition y(0) = 0. The solutions provided seem to be a bit mixed up, but let's go through the process of solving this differential equation:

1. This is a first-order linear differential equation. We can solve it using an integrating factor. The integrating factor is e^(∫p(t) dt), where p(t) is the coefficient of y in the differential equation. In this case, p(t) = 2, so the integrating factor is e^(2t).

2. Multiply every term in the differential equation by the integrating factor: e^(2t)y'(t) + 2e^(2t)y(t) = 2e^(2t).

3. The left-hand side of this equation is now the derivative of (e^(2t)y(t)) with respect to t. So, we can write the equation as (e^(2t)y(t))' = 2e^(2t).

4. Integrate both sides with respect to t: ∫(e^(2t)y(t))' dt = ∫2e^(2t) dt. This gives e^(2t)y(t) = e^(2t) + C, where C is the constant of integration.

5. Solve for y(t): y(t) = 1 + Ce^(-2t).

6. Apply the initial condition y(0) = 0 to find C: 0 = 1 + C, so C = -1.

7. Substitute C = -1 back into the equation for y(t) to get the final solution: y(t) = 1 - e^(-2t).

So, the correct solution is y(t) = 1 - e^(-2t).