Question

What can we conclude for the following linear homogeneous equation? t2y''+3ty'+y=0, t>0. y1=t is a solution. By the method of reduction of order, we can ALWAYS find another independent solution y2 satisfying W(y1,y2)≠ 0 y1=t−1 is a solution. By the method of reduction of order, we can ALWAYS find another independent solution y2 satisfying W(y1,y2)≠ 0 None of these y1=t is a solution. By the method of reduction of order, we can SOMETIMES find another independent solution y2 satisfying W(y1,y2)≠ 0 y1=t−1 is a solution. By the method of reduction of order, we can SOMETIMES find another independent solution y2 satisfying W(y1,y2)≠ 0

Answer 1

Required conclusion is that if
`y_1, y_2` satisfies given differential equation and wronskean is zero then they are considered as solution of that differential equation.

Explanation:

Given differential equation,

`t^2y''+3ty'+y=0`
`t>0\hfill (1)`

(i) To verify
`y_1(t)=t` is a solution or not we have to show,

`t^2y_(1)^('')+3ty_(1)^(')+y_1=0`

But,

`t^2y_(1)^('')+3ty_(1)^(')+y_1=(t^2* 0)=(3t* 1)+t=4t\\eq 0`

hence
`y_1` is not a solution of (1).

Now if
`y_2=t-1` is another solution where
`y_2(t)=t-1` then,

`t^2y_(2)^('')+3ty_(2)^(')+y_2=0`

But,

`t^2y_(2)^('')+3ty_(2)^(')+y_2=(t^2* 0)+(3t* 1)+t-1=4t-1\\eq 0`

so
`y_2` is not a solution of (1).

(ii) Rather the wronskean,

`W(y_1,y_2)=y_(1)y_(2)^(')-y_(2)y_(1)^(')=(t* 1)-((t-1)* 1)=t-t+1=1\\eq 0`

Hence it is conclude that if
`y_1, y_2` satisfies (i) along with condition (ii) that is wronskean zero, only then
`y_1, y_2` will consider as solution of (1).