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''=(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\frac{2{y_2}'+(3+t)y_2}{t^2}-y_2\frac{2{y_1}'+(3+t)y_1}{t^2}`

`\implies W'=(2W)/(t^2)`

which is another separable ODE; we have

`\frac{\mathrm dW}W=\frac2{t^2}\,\mathrm dt\implies \ln|W|=-\frac2t+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)}`