Computing Wronskian when functions are not given 
If $y_1,y_2$ are solutions of $y''+\sin(t)y'+e^ty=0$, and $y_1(0)=2,y_1'(0)=-3, y_2(0)=3, y_2'(0)=4$, compute $W[y_1,y_2](-5\pi)$.

I don't know how to calculate this wronskian without finding $y_1,y_2$ explicitly. Any idea?
 A: Hint: The Wronskian of two functions, as a function, also satisfies a certain ODE, see e.g. https://tutorial.math.lamar.edu/Classes/DE/Wronskian.aspx.
A: If $y_1$ and $y_2$ satisfy the second order, ordinary differential differential equation
$y'' + py' + qy = 0, \tag 1$
then their Wronskian
$W[y_1, y_2] = \det \begin{bmatrix} y_1 & y_2 \\ y_1' & y_2' \end{bmatrix} = y_1y_2' - y_2 y_1' \tag 2$
obeys the simple differential equation
$W' = -pW, \tag 3$
which may be seen as follows:  from (2),
$W' = y_1'y_2' + y_1y_2'' - y_2'y_1' - y_2y_1'' = y_1y_2'' - y_2y_1''; \tag 4$
from (1),
$y_1'' = -py_1' - qy_1, \; y_2'' = -py_2' - qy_2; \tag 5$
we substitute $y_1''$ and $y_2''$ from (5) into (4) and find
$W' = y_1(-(py_2' + qy_2)) - y_2(-(py_1' + qy_1)) = -py_1y_2' -qy_1y_2 + py_2y_1' + qy_1y_2$
$= -py_1y_2'+ py_2y_1' = -p(y_1y_2' - y_2y_1') =-pW, \tag 6$
the differential equation for $W$.  In the present case (1) takes the specific form
$y'' + (\sin t) y' + e^ty = 0, \tag 7$
so that
$p(t) = \sin t; \tag 8$
thus,
$W' = -(\sin t) W; \tag 9$
the solution to this equation is well-known to be
$W(t) = \displaystyle \exp  \left( -\int_{t_0}^t \sin s \; ds \right) W(t_0); \tag{10}$
since
$\displaystyle \int_{t_0}^t \sin s \; ds = \cos t_0 - \cos t, \tag{11}$
(10) becomes
$W(t) = \exp  ( \cos t - \cos t_0 ) W(t_0), \tag{12}$
which the reader may easily check by direct differentiation.
Taking
$t_0 = 0, \tag{13}$
(12) yields
$W(t) = \exp  ( \cos t - 1) W(t_0), \tag{14}$
and we may calculate $W(0)$ based upon the given values
$y_1(0)  = 2, \; y_1'(0) = -3, \; y_2(0) = 3, \; y_2'(0) = 4: \tag{15}$
$W(0) = y_1(0)y_2'(0) - y_1'(0)y_2(0) = 2 \cdot 4 - (-3) \cdot 3 = 8 + 9 = 17; \tag{16}$
now (14) yields
$W(t) = 17 \exp  ( \cos t - 1). \tag{17}$
Finally, we find
$W(-5\pi) = 17 \exp(\cos(-5\pi) - 1)$
$= 17 \exp (-1 -1) = 17\exp(-2) = \dfrac{17}{e^2} \approx 2.300699815. \tag{18}$
The reader will observe we haven't used the explicit form of either $y_1$ or $y_2$ to obtain this result.
