Answer: The verification is done below.
Step-by-step explanation: We are given to show that the function [tex]y=f(t)=2t+5t\ln t[/tex] is a solution to the following differential equation :
[tex]t^2y^{\prime\prime}-ty^\prime+y=0~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)[/tex]
If [tex]y=f(t)=2t+5t\ln t,[/tex] we have
[tex]y^\prime=\dfrac{d}{dt}f(t)=\dfrac{d}{dt}(2t+5t\ln t)=2\times1+5\ln t+5t\times\dfrac{1}{t}\\\\\\\Rightarrow y^{\prime}=7+5\ln t,\\\\\\y^{\prime\prime}=\dfrac{d}{dt}(7+5\ln t)=0+5\times\dfrac{1}{t}=\dfrac{5}{t}.[/tex]
Therefore, we get
[tex]L.H.S.\\\\=t^2y^{\prime\prime}-ty^\prime+y\\\\=t^2\times\dfrac{5}{t}-t(7+5\ln t)+(2t+5t \ln t)\\\\\\=5t-7t-5t\ln t+2t+5t\ln t\\\\=0\\\\=R.H.S.[/tex]
Thus, the function [tex]y=f(t)=2t+5t\ln t[/tex] is a solution to the given differential equation.
Hence showed.
