Final answer:
To solve the given first-order linear differential equations, we apply the method of integrating factors, multiplying through by an exponential function of an integral, and then integrating to find the general and particular solutions.
Step-by-step explanation:
To solve the differential equation dy/dt + y\(\cos(t)\) = 0, we can apply the method of integrating factors. The integrating factor here is \(e^{\int \cos(t) dt} = e^{\sin(t)}\). Multiplying both sides of the differential equation by this integrating factor, we have:
\(e^{\sin(t)}\frac{dy}{dt} + e^{\sin(t)}y\cos(t) = 0\)
Which simplifies to:
\(\frac{d}{dt}(ye^{\sin(t)}) = 0\)
Integrating both sides with respect to \(t\) gives:
\(ye^{\sin(t)} = C\)
Where \(C\) is the constant of integration. The general solution to the differential equation is:
\(y = Ce^{-\sin(t)}\).
For the initial value problem (IVP) dy/dt - 2ty = t, with the given conditions, we can also use an integrating factor. The integrating factor in this case is \(e^{-t^2}\). Multiplying both sides by the integrating factor gives:
\(e^{-t^2}\frac{dy}{dt} - 2te^{-t^2}y = te^{-t^2}\)
Which simplifies to:
\(\frac{d}{dt}(ye^{-t^2}) = te^{-t^2}\)
Integrating both sides with respect to \(t\) and applying the initial conditions will yield the particular solution.