Question

If the differential equation t 2y 00 − 2y 0 + (3 + t)y = 0 has y1(t) and y2(t) as a fundamental set of solutions and if W[y1, y2](2) = 3, find the value of W[y1, y2](6) ?

Answer 1

In the ODE, solve for $y''$:

$t^2y''-2y'+(3+t)y=0\implies y''=\dfrac{2y'+(3+t)y}{t^2}$

The Wronskian is then

$W=\begin{vmatrix}y_1&y_2\\{y_1}'&{y_2}'\end{vmatrix}=y_1{y_2}'-{y_1}'y_2$

Differentiating the Wronskian gives

$W'=({y_1}'{y_2}'+y_1{y_2}'')-({y_1}''y_2+{y_1}'{y_2}')=y_1{y_2}''-{y_1}''y_2$

Substitute $y_1,y_2$ into the equation for $y''$, then substitute ${y_1}'',{y_2}''$ into $W'$:

$W'=y_1\dfrac{2{y_2}'+(3+t)y_2}{t^2}-y_2\dfrac{2{y_1}'+(3+t)y_1}{t^2}$

$\implies W'=\dfrac{2W}{t^2}$

which is another separable ODE; we have

$\dfrac{\mathrm dW}W=\dfrac2{t^2}\,\mathrm dt\implies \ln|W|=-\dfrac2t+C\implies W=Ce^{-2/t}$

Given that $W(y_1,y_2)(2)=3$, we find

$3=Ce^{-2/2}\implies C=3e$

so that

$W(y_1,y_2)(t)=3e^{1-2/t}$

and so

$W(y_1,y_2)(6)=\boxed{3e^{2/3}}$