Question

Verify that the given functions y1 and y2 satisfy the corresponding homogeneous equation; then find a particular solution of the given nonhomogeneous equation. t2y'' − 2y = 5t2 − 1, t > 0, y1(t) = t2, y2(t) = t−1

Answer 1

`y_1, y_2` satisfies given homogenous solution and the particular solution is
`(5)/(t^4)+(1)/(2t^2)-(5)/(2)`.

Explanation:

Given homogeneous equation is,

`t^2y''-2y=5t^2-1` where t>0 subject to
`y_1(t)=t^2, y_2(t)=(1)/(t)\hfill (1)`

To verify,

• wheather
  `y_1, y_2` satisfies given homogenous equation, then both will satisfy
  `t^2y''-2y=0` that is,

`t^2y_(1)^('')-2y_1=t^2* 2-2* t^2=0`

`t^2y_(2)^('')-2y_2=t^2* 2t^(-3)-2* t^(-1)=2t^(-1)-2t^(-1)=0`

Thus
`y_1` and
`y_2` satisfies (1).

Now the wronskean,

`W(y_1, y_2)(x)=y_1y_(2)^(')-y_(2)y_(1)^(')=-t^2t^(-2)-2tt^(-1)=-3\\eq 0`

Thus
`y_1` and
`y_2` are solution of (1) .

• Particular solution,

`Y(t)=(1)/(t^2D^2-2)(5t^2-1)` where
`D\equiv (\partial )/(\partial t)`

`=-(1)/(2t^2)(1)/(1-(D^2)/(t^2))()5t^2-1`

`=-(1)/(2t^2)(1-(D^2)/(t^2))^(-1)(5t^2-1)`

`=-(1)/(2t^2)(1-(D^2)/(t^2)+........)(5t^2-1)`

`=-(1)/(2t^2)(5t^2-1-(10)/(t^2))` sincce
`D^2(5t^2-1)=10`

`=(5)/(t^4)+(1)/(2t^2)-(5)/(2)`

Hence the result.