Question

Show that assuming y=e^lambda(t) we are led to the equation... a(lambda^2)+b(lambda)+c=0

Answer 1

I'm assuming this is for Diff Eq?

Suppose you have the ODE \[ay''+by'+cy=0\]
Assume \(y=e^{\lambda t}\), which gives you
\[y'=\lambda e^{\lambda t}\text{ and }y''=\lambda^2e^{\lambda t}\]
The ODE then changes to
\[a\left(\lambda^2e^{\lambda t}\right)+b\left(\lambda e^{\lambda t}\right)+c\left(e^{\lambda t}\right)=0\\
e^{\lambda t}\left(a\lambda^2+b\lambda +c\right)=0\]
\(e^{\lambda t}>0\text{ for all }t,\) so you can divide both sides by it, leaving you with
\[a\lambda^2+b\lambda +c=0\]

Answer 2

Yes you are completely correct, it is for Differential Equation. :) Thanks for the help! :D