Answer:
[tex]L(tsinn^2 t)=\dfrac{2sn^2}{s^2+n^4}[/tex]
Step-by-step explanation:
Given
[tex]f(t)=tsin n^2 t[/tex]
Here n is constant because f(t) is a function of t so n will be treated as constant.
We know that
[tex]L(t^nf(t))=(-1)^n\dfrac{d^nF(S)}{dS^n}[/tex]
Here given that f(t)=sin n^2t
So [tex]F(s)=\dfrac{ n^2}{S^2+ n^4}[/tex]
Find the derivative of F(s)
[tex]\dfrac{dF(S)}{dS}=\dfrac{-2sn^2}{s^2+n^4}[/tex]
So
[tex]L(tsinn^2 t)=\dfrac{2sn^2}{s^2+n^4}[/tex]
