Differential equations

[blizzard750]

y1(t) = t^(2)+5t, y2(t) = t^(2)-5t

I know that these functions are linearly independent because they are not scalar multiples, but every time that i do the Wronskian i get 0.

[t^(2) + 5t t^(2)-5t]
[2t+5 2t -5]

(2t^(3) -25t) - (2t^3-25t) = 0 my instructor some how gets 20t please tell me how

[lurflurf]

Did you write that down wrong?
It should be 10t^2
(t^(2) + 5t)(t^(2) - 5t)'=(t^(2) + 5t)(2t -5)=2t^3+5t^2-25!=(2t^(3) -25t)
(t^(2) - 5t)(t^(2) + 5t)=(t^(2) - 5t)(2t +5)=2t^3-5t^2-25!=(2t^(3) -25t)
W=(2t^3+5t^2-25)-(2t^3-5t^2-25)
and one way a person would get 20t is if he/she took
(t^(2) + 5t)'=2t