Question

Find the solution to dy/dt = 7 y

satisfying
y(9) = 5

Answer 1

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

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.

Answer 2

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

Learn more:https://brainly.com/question/12685592