dy/dt = y + 2
y(0) = 3
find y(2)
It's mainly the combination of dy/dt and y in the equation that's causing my confusion. Does this mean that when I take the antiderivative, I need to substitute so that it is with respect to y?
$\displaystyle \frac{dy}{dt}=y+2$
then
$\displaystyle \int\frac{dy}{y+2}=\int dt$
after moving differentials and integrating. Taking the antiderivative,
$\displaystyle \ln(y+2)=t+C$ which, after solving for y, leads to
$\displaystyle y=Ce^t - 2.$ Plugging in y(0)=3, $\displaystyle 3=Ce^0 - 2 = C - 2$ and after solving for C, C = 5.
So our specific solution is $\displaystyle y(t)=5e^t-2$. Feel free to verify this by differentiating and using initial conditions. Finding y(2),
$\displaystyle y(2)=5e^2 - 2$.
(As a side note, go Ovie!)