Computing Wronskian when functions are not given [closed]

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?

Solution 1:

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.

Solution 2:

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.