Answer to Question #195223 in Differential Equations for Sourav

Question

Do the functions 

y1(t)=√t and y2(t)=1/t

form a fundamental set of solutions of the 

equation 2t²y"+3ty'-y=0

, on the interval 

0 < t <infinity

Justify your answer.

Answer 1

$y_1(t)={\sqrt t}$

$y_2(t)=\dfrac{1}{t}$

Given equation is -

$2t^{2}y''$ $+$ $3ty'-y=0$

Now to check $y_1(t)$ and $y_2(t)$ are set of solutions of this equation , we will check this by the help of wronskian Abel's theorem of method of solving differential equation .

$y_1(t)={\sqrt t}$

$y_1'(t)=\dfrac{1}{2{\sqrt t}}$

$y_1''(t)=$

$y_2(t)=\dfrac{1}{t}$

$y_2'(t)=-\dfrac{1}{t^{2}}$

$y_2''(t)=2t^{-3}$

Now for checking that these two are the solution of this differential equation , we compute wronskian method of solving differential equation ,which is -

$\begin{vmatrix} y_1& y_2 \\ y_1' & y_2' \end{vmatrix}$ =$y_1y_2'-y_2y_1'=\dfrac{3}{2}t^{\dfrac{-3}{2}}$

Notice than wronskian is zero at t=0 but non zero at t=1 . By the above corollary , $y_1$ and $y_2 \$ cannot be both be solution in the respective interval $0<t<{\infty}$ .

Now to see whether given functions satisfy this diiferential equation we can also check it by expressing in terms of determinant -

$\begin{vmatrix} y_1 & y_2 &y\\ y_1'& y_2' &y' \\ y1'' &y_2'' & y'' \end{vmatrix}$ $=0$

Now putting the values in determinant , and expanding along the 1st column , we get -

$\begin{vmatrix} {\sqrt t} & \dfrac{1}{t} &y\\ \dfrac{1}{2{\sqrt t}}& \dfrac{-1}{t^{2}} &y' \\ \dfrac{-1}{4}t^{\dfrac{-3}{2}} &2t^{-3} & y'' \end{vmatrix}$ $=0$

Now expanding this determinant , we get -

$=2y''t^{2}-5ty'-5y=0$

which is form differential equation , so the given roots are not equation of the given equation .hence proved .