Question

Suppose y(t) = 3e^-3t is a solution of the initial value problem y' + ky = 0, y(0) = y0. What are the constants k and y0?

Answer 1

If \(y(t) = 3e^{-3t}\) what is \(y'\)?

Answer 2

3/e^3?

Answer 3

No, \[y' = \frac{d}{dt}[{3e^{-3t}}] = -3*3e^{-3t} = -9e^{-3t}\]

Answer 4

That gives us
\[y' + ky = 0\]\[-9e^{-3t} + 3ke^{-3t} = 0\]\[y(0) = y_0 = -9e^{-3(0)} +3ke^{-3(0)} = -9(1)+3k(1) = 3k-9\]But we also know that\[-9e^{-3(0)}+3ke^{-3(0)} = 0\rightarrow -9+3k=0\]\[k=3\]and you can find \(y_0\) from there...