Respuesta :
Answer:
Required conclusion is that if [tex]y_1, y_2[/tex] satisfies given differential equation and wronskean is zero then they are considered as solution of that differential equation.
Step-by-step explanation:
Given differential equation,
[tex]t^2y''+3ty'+y=0[/tex] [tex] t>0\hfill (1)[/tex]
(i) To verify [tex]y_1(t)=t[/tex] is a solution or not we have to show,
[tex]t^2y_{1}^{''}+3ty_{1}^{'}+y_1=0[/tex]
But,
[tex]t^2y_{1}^{''}+3ty_{1}^{'}+y_1=(t^2\times 0)=(3t\times 1)+t=4t\neq 0[/tex]
hence [tex]y_1[/tex] is not a solution of (1).
Now if [tex]y_2=t-1[/tex] is another solution where [tex]y_2(t)=t-1[/tex] then,
[tex]t^2y_{2}^{''}+3ty_{2}^{'}+y_2=0[/tex]
But,
[tex]t^2y_{2}^{''}+3ty_{2}^{'}+y_2=(t^2\times 0)+(3t\times 1)+t-1=4t-1\neq 0[/tex]
so [tex]y_2[/tex] is not a solution of (1).
(ii) Rather the wronskean,
[tex]W(y_1,y_2)=y_{1}y_{2}^{'}-y_{2}y_{1}^{'}=(t\times 1)-((t-1)\times 1)=t-t+1=1\neq 0[/tex]
Hence it is conclude that if [tex]y_1, y_2[/tex] satisfies (i) along with condition (ii) that is wronskean zero, only then [tex]y_1, y_2[/tex] will consider as solution of (1).
