Final answer:
To solve the differential equation 2y' - y = e^(t/3) with initial condition y(0) = 2, we can use the method of integrating factors. The particular solution is y = 2e^(t/2) - 3e^(-t/6) + 2.
Step-by-step explanation:
To solve the given equation 2y' - y = e^(t/3), we can use the method of integrating factors. First, rewrite the equation in the form y' - (1/2)y = (1/2)e^(t/3). The integrating factor is e^(int(-1/2) dt) = e^(-t/2). Multiply both sides of the equation by e^(-t/2) to obtain e^(-t/2)y' - (1/2)e^(-t/2)y = (1/2)e^(t/3)e^(-t/2) = (1/2)e^(t/6).
The left side of the equation can be rewritten as d/dt (e^(-t/2)y) using the product rule. So, d/dt (e^(-t/2)y) = (1/2)e^(t/6), which can be integrated to find e^(-t/2)y. Integrate both sides and solve for y to obtain y = 2e^(t/2) - 3e^(-t/6) + C.
Using the initial condition y(0) = 2, we can substitute t = 0 and y = 2 into the equation to find the value of the constant C. Therefore, 2 = 2e^(0/2) - 3e^(-0/6) + C, which simplifies to C = 2. Hence, the particular solution to the equation with the initial condition y(0) = 2 is y = 2e^(t/2) - 3e^(-t/6) + 2.