# Math Help - Help with exam prep

1. ## Help with exam prep

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?

2. $\frac{dy}{dt}=y+2$

then

$\int\frac{dy}{y+2}=\int dt$

after moving differentials and integrating. Taking the antiderivative,

$\ln(y+2)=t+C$ which, after solving for y, leads to

$y=Ce^t - 2.$ Plugging in y(0)=3, $3=Ce^0 - 2 = C - 2$ and after solving for C, C = 5.

So our specific solution is $y(t)=5e^t-2$. Feel free to verify this by differentiating and using initial conditions. Finding y(2),

$y(2)=5e^2 - 2$.

(As a side note, go Ovie!)