Question

Verify that y1(t) =t2 and y2(t) =t−1 are two solutions of the differential equation t2y−2y=0 for t > 0.Then show that y=c1t2 +c2t−1 is also a solution of this equation for any c1 and c2.

Answer 1

Answer:

Step-by-step explanation:

Consider first

$y_1(t) = t^2$

We differentiate this two times to get

$y_1'(t) = 2t\\y_1"(t) =2$

Substitute in the given equation

$t^2 (2) -2(t^2 =0$

Hence satisfied

Consider II equation

$y_2(t) = t^{-1} \\y_2'(t) = - t^{-2}\\y_2"(t) = -2 t^{-3}$

Substitute in the given equation to get

$t^2 (-2 t^{-3})+2 t^{-1} = 0$

Hence satisfied

Together if we have

$y = c_1 t^2 +c_2 t^{-1}$

being linear combination of two solutions

automatically this also will satisfiy the DE Or

this is a solution to the given DE