Question 3: Linear Homogeneous Equations

If the two solutions to a linear second-order homogeneous equation are $- 4 ⅇ^{- 5 t} \cos (t)$ and $- 2 \sin (4 t)$ , find the Wronskian matrix $W (t)$ .

[1] [0]

$(\begin{array}{c} - 4 ⅇ^{- 5 t} \cos (t) & - 2 \sin (4 t) \\ 4 ⅇ^{- 5 t} \sin (t) + 20 ⅇ^{- 5 t} \cos (t) & - 8 \cos (4 t) \end{array})$

Step 1 of 3

For a linear second-order homogeneous equation, the Wronskian matrix is:

$W (t) =$ $(\begin{matrix} f_{1} (t) & f_{2} (t) & \dots & f_{n} (t) \\ {f_{1}}^{'} (t) & {f_{2}}^{'} (t) & \dots & {f_{n}}^{'} (t) \\ ⋮ & ⋮ & ⋱ & ⋮ \\ {f_{1}}^{n - 1} (t) & {f_{2}}^{n - 1} (t) & \dots & {f_{n}}^{n - 1} (t) \end{matrix})$

where $f_{1} (t)$ , $f_{2} (t)$ , ..., $f_{n} (t)$ are solutions to the linear second-order homogeneous equation.

Step 2 of 3

Therefore, the Wronskian matrix for the linear second-order homogeneous equation with solutions $- 4 ⅇ^{- 5 t} \cos (t)$ and $- 2 \sin (4 t)$ is:

$W (t) =$ $(\begin{matrix} - 4 ⅇ^{- 5 t} \cos (t) & - 2 \sin (4 t) \\ \frac{d}{d t} - 4 ⅇ^{- 5 t} \cos (t) & \frac{d}{d t} - 2 \sin (4 t) \end{matrix})$

Step 3 of 3

After computing the derivatives, we have:

$(\begin{array}{c} - 4 ⅇ^{- 5 t} \cos (t) & - 2 \sin (4 t) \\ 20 ⅇ^{- 5 t} \cos (t) + 4 ⅇ^{- 5 t} \sin (t) & - 8 \cos (4 t) \end{array})$

Since we can also assume $f_{1} (t) =$ $- 2 \sin (4 t)$ and $f_{2} (t) =$ $- 4 ⅇ^{- 5 t} \cos (t)$ , the answer could also be:

$(\begin{array}{c} - 2 \sin (4 t) & - 4 ⅇ^{- 5 t} \cos (t) \\ - 8 \cos (4 t) & 4 ⅇ^{- 5 t} \sin (t) + 20 ⅇ^{- 5 t} \cos (t) \end{array})$