Question

verify that each given function is a solution of the

differential equation.

9.t^2y′′\:+\:5ty′\:+\:4y\:=\:0;\:y\left(t\right)\:=\:t^{−2}

Answer 1

Let us verify that the function $\:y\left(t\right)\:=\:t^{−2}$ is a solution of the differential equation $t^2y''\:+\:5ty'\:+\:4y\:=\:0.$ Taking into account that $y'(t)=-2t^{-3}, \ y''(t)=6t^{-4},$ and

$t^2y''(t)+\:5ty'(t)+4y=t^2(6t^{-4})+\:5t(-2t^{-3})+4t^{-2}=6t^{-2}-10t^{-2}+4t^{-2}=0,$

we conclude that the function $\:y\left(t\right)\:=\:t^{−2}$ is a solution of the differential equation $t^2y''\:+\:5ty'\:+\:4y\:=\:0.$