Question

Show that the function: y=f(t)= 2t +5tln(t) is a solution to the differential equation: t'y"-ty'+ y = 0.

Answer 1

Answer: The verification is done below.

Step-by-step explanation: We are given to show that the function
`y=f(t)=2t+5t\ln t` is a solution to the following differential equation :

`t^2y^(\prime\prime)-ty^\prime+y=0~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)`

If
`y=f(t)=2t+5t\ln t,` we have

`y^\prime=(d)/(dt)f(t)=(d)/(dt)(2t+5t\ln t)=2*1+5\ln t+5t*(1)/(t)\\\\\\\Rightarrow y^(\prime)=7+5\ln t,\\\\\\y^(\prime\prime)=(d)/(dt)(7+5\ln t)=0+5*(1)/(t)=(5)/(t).`

Therefore, we get

`L.H.S.\\\\=t^2y^(\prime\prime)-ty^\prime+y\\\\=t^2*(5)/(t)-t(7+5\ln t)+(2t+5t \ln t)\\\\\\=5t-7t-5t\ln t+2t+5t\ln t\\\\=0\\\\=R.H.S.`

Thus, the function
`y=f(t)=2t+5t\ln t` is a solution to the given differential equation.

Hence showed.