Verified Answer :

Zepdrix

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

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,

\(\rm y=A e^{7t}\qquad\qquad A e0\)

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.

yoodyannapolis

The solution is \(y(t)=5{e^{7({t-9})}\)

We have given differential equation as

\(\frac{dy}{dt} = 7 y\) ,y(9)=5

This is a separable differential equation can be solved as

\(\frac{dy}{7y} = dt\)

Integrate both sides

\(\int\frac{dy}{7y} =\int dt\\\frac{1}{7} logy=t+c\\logy=7t+C\\y=Ae^{7t}\)

We have y(9)=5

\(5=Ae^{63}\\A=\frac{5}{e^{63}}\)

So,

\(y=\frac{5}{e^{63}} e^{7t}\\y(t)=5{e^{7t-63}\\\)

\(y(t)=5{e^{7({t-9})}\)

Therefore the solution is \(y(t)=5{e^{7({t-9})}\)

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