Answer:
[tex]y=-A+Ce^{kt}[/tex]
Step-by-step explanation:
Let A and k be positive constants
[tex]\frac{dy}{dt} =k(y+A)[/tex]
Here k and A are constant.
[tex]\frac{dy}{dt} =k(y+A)[/tex]
multiply both sides by dt
[tex]dy=k(y+A)dt[/tex]
[tex]\frac{dy}{y+A} =kdt[/tex]
Integrate on both sides
[tex]ln|y+A|=kt+C[/tex]
Now solve for y. convert it into exponential form
[tex]y+A=e^{kt+C}[/tex]
[tex]y+A=e^{kt}e^C[/tex]
[tex]y+A=Ce^{kt}[/tex]
Subtract A from both sides
[tex]y=-A+Ce^{kt}[/tex]
