Question

I have = \frac{ t^{2} }{9} " alt="f(t)= \frac{ t^{2} }{9} " align="absmiddle" class="latex-formula"> and $g(t)= e^{ \frac{- ln(t)}{4} }$
Given that $k(t)=f(t)+cg(t)$ and k(1)=4, what is the value of c?

Answer 1

$\bf \begin{cases} f(t)=\cfrac{t^2}{9}\\\\ g(t)=e^{\cfrac{}{}\frac{-ln(t)}{4}}\\\\ k(t)=f(t)+c\cdot g(t) \end{cases}\implies k(t)=\cfrac{t^2}{9}+C\cdot e^{\cfrac{}{}\frac{-ln(t)}{4}} \\\\\\ k(1)=4\implies \begin{cases} k=4\\ t=1 \end{cases}\implies 4=\cfrac{1^2}{9}+C\cdot e^{\cfrac{}{}\frac{-ln(1)}{4}} \\\\\\ 4=\cfrac{1}{9}+C\cdot e^{\frac{0}{4}}\implies 4=\cfrac{1}{9}+C\cdot 1\implies 4-\cfrac{1}{9}=C \\\\\\ \cfrac{35}{9}=C$