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

Answer:

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.

Step-by-step 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,

$t^2y_{1}^{''}+3ty_{1}^{'}+y_1=(t^2\times 0)=(3t\times 1)+t=4t\neq 0$

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,

$t^2y_{2}^{''}+3ty_{2}^{'}+y_2=(t^2\times 0)+(3t\times 1)+t-1=4t-1\neq 0$

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

(ii) Rather the wronskean,

$W(y_1,y_2)=y_{1}y_{2}^{'}-y_{2}y_{1}^{'}=(t\times 1)-((t-1)\times 1)=t-t+1=1\neq 0$

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).