Final answer:
The first order ODE 9(dv/dt) + 2v(t) = -9, with the initial condition v(0) = 4, has a solution in the form v(t) = Ae^(-t/τ) + B. By applying integrating factors and matching coefficients, we find that τ = 9/2, A = 4, and B = -1/2.
Step-by-step explanation:
To solve the given first order ordinary differential equation (ODE) 9(dv/dt) + 2v(t) = -9, with the initial condition v(0) = 4, we use the standard approach for first order linear ODEs.
First, we simplify the equation and identify the integrating factor. The ODE can be rewritten as:
dv/dt + (2/9)v(t) = -1
The integrating factor, μ(t), is obtained by exponentiating the integral of the coefficient of v(t), which in this case is (2/9)t:
μ(t) = e^(2/9)t
Multiplying the entire ODE by the integrating factor, we get the left side as the derivative of the product of μ(t) and v(t):
d(μ(t)*v(t))/dt = μ(t)*(-1)
Integrating both sides with respect to t gives:
μ(t)*v(t) = -μ(t)/a + C
We solve for C using the initial condition v(0) = 4:
C = μ(0)*4 = e^(0)*4 = 4
Now we find the general solution v(t), which is:
v(t) = (1/μ(t))(C - μ(t)/a)
And by finding the values from the integrating factor and the constant C:
v(t) = (e^(-2/9)t)*(4 - (e^(2/9)t)/a)
Now we match this general solution with the given form v(t) = Ae^(-t/τ) + B to identify A, B, and τ. Equating coefficients, we can see that τ = 9/2, A = 4, and B equals the limit of the function as t approaches infinity, which equals -1/2.