Consider the initial value problem $\displaystyle{\frac{dy}{dt} = \frac{1}{2}(y+1)}$, where $y(0)=0$.

Use the differential equation to find the slope of the tangent line to the solution curve $y=y(t)$ at $t = 0$.
Slope =

Use the given initial value to find the equation of the tangent line to the solution curve at $t = 0$.
$y =$

Use the equation of the tangent line that you just found to approximate $y(2)$, the value of the solution at $t=2$.
$y(2) \approx$

Assuming that your approximation for $y(2)$ is the actual value of $y(2)$, use the differential equation to find the slope of the tangent line to $y=y(t)$ at $t = 2$.
Slope =

Use the slope that you just found to find an approximate equation of the tangent line to the solution curve at $t = 2$.
$y \approx$

Use the equation of the tangent line that you just found to approximate $y(4)$, the value of the solution at $t=4$.
$y(4) \approx$