Here are some initial value problems with obvious solutions, as discussed in class. In all cases the solutions are functions of $x$. All letters other than $y$ and $x$ denote constants.

The solution of $$y' = ky,\quad y(0) = A$$ is

$y(x) =$ .

The solution of $$y'' = k^2y,\quad y(1) = y(-1) = A$$ is

$y(x) =$ .

The solution of $$y'' = k^2y,\quad y(1) = -y(-1) = A$$ is

$y(x) =$ .

The solution of $$y'' = -k^2y,\quad y(0) = 1, \quad y'(0) = 0$$ is

$y(x) =$ .

The solution of $$y'' = -k^2y,\quad y(0) = 0, \quad y'(0) = 1$$ is

$y(x) =$ .

The solution of $$y'' = -k^2y,\quad y(0) = A, \quad y'(0) = B$$ is

$y(x) =$ .