Final answer:
Therefore, the critical value a0 is a0 = 2e^(-PI/4). If a < a0, the solution is stable and approaches zero as t approaches infinity. If a > a0, the solution is unstable and approaches a nonzero constant as t approaches infinity.
Step-by-step explanation:
To solve the initial value problem 3y’-2y=e^((-PIt)/2) and y(0)=a, we can use the method of integrating factors. The integrating factor for this differential equation is e^(3t/2). Multiplying both sides of the equation by this factor, we get:
e^(3t/2) * 3y’ - 2e^(3t/2) * y = e^(t/2 - PI/4)
We can rewrite this equation as:
(d/dt)(e^(3t/2) * y) = e^(t/2 - PI/4)
Integrating both sides concerning t, we get:
e^(3t/2) * y = 2e^(t/2 - PI/4) + C
where C is the constant of integration. Solving for y, we get:
y = (2e^(t/2 - PI/4) + C) / e^(3t/2)
Using the initial condition y(0) = a, we get:
a = (2e^(-PI/4) + C)/1
Solving for C, we get:
C = a - 2e^(-PI/4)
Therefore, the solution to the initial value problem is:
y = (2e^(t/2 - PI/4) + a - 2e^(-PI/4))/e^(3t/2)
To find the critical value a0, we need to determine the value of a for which the transition from one type of long-run behavior to another occurs.
The denominator of the solution is e^(3t/2), which approaches infinity as t approaches infinity.
Therefore, the long-run behavior of the solution is determined by the numerator. If the numerator approaches zero as t approaches infinity, the solution approaches zero and is said to be stable.
If the numerator approaches a nonzero constant as t approaches infinity, the solution approaches that constant and is said to be unstable.
Setting the numerator equal to zero, we get:
2e^(t/2 - PI/4) + a - 2e^(-PI/4) = 0
Solving for a, we get:
a = 2e^(-t/2 + PI/4) + 2e^(-PI/4)
Therefore, the critical value a0 is:
a0 = 2e^(-PI/4)
If a < a0, the solution is stable and approaches zero as t approaches infinity. If a > a0, the solution is unstable and approaches a nonzero constant as t approaches infinity.