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?

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- Dec 9th 2008, 02:28 PMOvechkinHelp 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? - Dec 9th 2008, 04:42 PMZiaris
$\displaystyle \frac{dy}{dt}=y+2$

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!)