The given differential equation is:
y''(t) + (t+4)y'(t) + e^(-3t)y(t) = 0
Suppose that y1(t) and y2(t) are two solutions to this differential equation. Then the Wronskian of these solutions is given by:
W(y1, y2)(t) = y1(t)y2'(t) - y1'(t)y2(t)
To find the Wronskian for any t, we can differentiate this expression with respect to t:
d/dt [W(y1, y2)(t)] = d/dt [y1(t)y2'(t) - y1'(t)y2(t)]
Using the product rule and the fact that y1(t) and y2(t) are solutions to the differential equation, we get:
d/dt [W(y1, y2)(t)] = y1(t)y2''(t) + y1'(t)y2'(t) - y1''(t)y2(t) - y1'(t)y2'(t)
Simplifying this expression and using the fact that y1(t) and y2(t) are solutions to the differential equation, we get:
d/dt [W(y1, y2)(t)] = - (t+4) [y1(t)y2'(t) - y1'(t)y2(t)]
Now, we can use the initial condition that W(y1, y2)(0) = 20 to solve for the Wronskian at any t. Specifically, we have:
W(y1, y2)(t) = W(y1, y2)(0) e^(-int(t+4) dt)
Integrating the factor e^(-int(t+4) dt) with respect to t, we get:
W(y1, y2)(t) = 20 e^(-1/2(t+4)^2)
Therefore, the Wronskian of y1(t) and y2(t) for any t is given by:
W(y1, y2)(t) = 20 e^(-1/2(t+4)^2)