verify that y1=t^2 and y2=t^-1 are the solutions for t^2y"-2y=0??

Question

verify that y1=t^2 and y2=t^-1 are the solutions for t^2y"-2y=0??
how do you do this prplem??

Answer 1

substitute the given values of y1 and y2 into the equation, then simplify

y1 = t^2
y1' = 2t
y1'' = 2
t^2 y1'' - 2 y1 = t^2 * 2 - 2 * t^2 = 0

y2 = t^(-1) = 1/t
y2' = -t^(-2) = -1/t^2
y2'' = 2 t^(-3) = 2/t^3
t^2 y2'' - 2 y2 = t^2 * 2/t^3 - 2*1/t = 2/t - 2/t = 0

Answer 2

Verify usually means substitute the solutions into the equation and make sure they check.

If they actually want a solution, well, that's slightly different. This is what's refered to as a homogenous second order differential equation with non constant coefficents. I'd suggest looking this one up in a mathematical methods book as it's somewhat tricky to solve. (IE: tricky enough that me beating on it for a few minutes got me an equation with the indp variable, dep variable, an integral involving both, and a derivative involving both!)

Answer 3

y1 = t^2
y'1 = 2t
y''1 = 2

t^2 * y'' - 2y = t^2 * 2 - 2 * t^2 = 0

y2 = t^-1
y'2 = -t^-2
y''2 = -2t^-3

t^2 * y'' - 2y = t^2 * -2t^-3 - 2*(t^-1) = 2*t^-1 - 2^t^-1 = 0

Answer 4

hmmm... is the 2y" representing two y double prime like in a differential equation? need a bit more info.

Answer 5

this is supposed to be your homework right.....dont cheat