Answer:
[tex]v=ce^{-\frac{t}{8}}+8[/tex]
Step-by-step explanation:
We have the differential equation:
[tex]\frac{dv}{dt}=1-\frac{v}{8}\\\\\frac{dv}{dt}+\frac{v}{8}=1\\\\v'+\frac{v}{8}=1\\[/tex]
This is linear differential equation. We conclude that are
[tex]p(t)=\frac{1}{8}\\q(t)=1[/tex]
We have the formula
[tex]v=e^{-\int p(t)\, dt}\left(c+\int q(t)\cdot e^{\int p(t)\, dt}\right)[/tex]
Now, we calculate the solution of the given differential equation:
[tex]v=e^{-\int \frac{1}{8} \, dt}\left(c+\int 1\cdot e^{\int \frac{1}{8} \, dt}\, dt\right)\\\\v=e^{-\frac{t}{8}}\left(c+\int e^{\frac{t}{8}}\, dt\right)\\\\v=e^{-\frac{t}{8}}\left(c+8e^{\frac{t}{8}}\right)\\v=ce^{-\frac{t}{8}}+8[/tex]
We use the site geogebra.org to sketch the solutions with the initial conditions v(0)= 5, 8, and 15.
