\( \rm \dfrac{dy}{dt}=7y,\qquad\qquad\qquad y(9)=5 \)
I’m not sure what methods you’ve learned up to this point but one option is to apply separation of variables:
\( \rm \dfrac{dy}{y}=7dt \)
and then integrate from there,
\( \rm ln|y|=7t+c \)
exponentiate to isolate y,
\( \rm |y|=e^{7t+c} \)
Apply exponent rule,
\( \rm |y|=e^c e^{7t} \)
rename this e^c as some new constant, perhaps A,
\( \rm \pm y=A e^{7t}\qquad\qquad A>0 \)
This A can only be positive, non-zero, but absorbing the plus/minus fixes that restriction,
\( \rm y=A e^{7t}\qquad\qquad A\ne0 \)
Use your initial information to solve for this unknown value A,
\( \rm y(9)=A e^{7\cdot9}=5 \)
solving for A, dividing by the exponential,
\( \rm A=5e^{-63} \)
So we get a final result of
\( \rm y(t)=5e^{-63}e^{7t} \)
apply exponent rule again to get a better looking answer, and factor,
\( \rm y(t)=5e^{7(t-9)} \)
Lemme know if too confusing.