Question

Suppose that yi (t) and y2 (t) are solutions to the following differential equation such that the Wronskian (yl. y2) (to=1) = 20. Find the Wronskian (y1, y2) (t) for any t. [Hint: see proof at top of page 143] dy(t) +e-3ty(t) = 0 dt d'y(t) (t + 4) dt2

Answer 1

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)