Answer:
Step-by-step explanation:
Consider first
[tex]y_1(t) = t^2\\[/tex]
We differentiate this two times to get
[tex]y_1'(t) = 2t\\y_1"(t) =2[/tex]
Substitute in the given equation
[tex]t^2 (2) -2(t^2 =0[/tex]
Hence satisfied
Consider II equation
[tex]y_2(t) = t^{-1} \\y_2'(t) = - t^{-2}\\y_2"(t) = -2 t^{-3}[/tex]
Substitute in the given equation to get
[tex]t^2 (-2 t^{-3})+2 t^{-1} = 0[/tex]
Hence satisfied
Together if we have
[tex]y = c_1 t^2 +c_2 t^{-1}[/tex]
being linear combination of two solutions
automatically this also will satisfiy the DE Or
this is a solution to the given DE
