satisfying

y(9) = 5

\( \rm \dfrac{dy}{dt}=7y,\qquad\qquad\qquad y(9)=5 \)

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.