Question

Solve the initial value problem. y' + 3y = 1, y(0) = 1 y = _____________ .

Answer 1

The solution to the initial value problem y' + 3y = 1 with y(0) = 1 is found by using an integrating factor, leading to the solution y = 1/3 + (2/3)e^{-3t}.

Step-by-step explanation:

To solve the initial value problem y' + 3y = 1, with the initial condition y(0) = 1, we will use the method of integrating factors.

1. Firstly, observe that the differential equation is linear and not homogeneous. We will look for an integrating factor, μ(t), such that multiplying the entire equation by μ(t) will make the left-hand side a product rule derivative, μ(t)y(t). The integrating factor is found using the formula μ(t) = e^{∫ 3dt} = e^{3t}.
2. Multiply both sides of the original differential equation by the integrating factor to get e^{3t}y' + 3e^{3t}y = e^{3t}.
3. Now, the left-hand side is the derivative of e^{3t}y with respect to t, so integrate both sides with respect to t to obtain e^{3t}y = ∫ e^{3t}dt = ⅓ e^{3t} + C, where C is the constant of integration.
4. Apply the initial condition y(0) = 1 to find C: e^{3·0}y(0) = 1 → 1·1 = ⅓ + C, which gives C = ⅓.
5. Now, solve for y by dividing both sides of the equation by e^{3t}: y = ⅓ + ⅓e^{-3t}.