Is $y'' = (57 + \sin t )y$ a stiff problem? Performing the change of variable $y_1 \to y$, $y_1' \to y_2$ gives an equivalent first order system $$y_1' = y_2$$ $$y_2'=(57+\sin t)y_1$$
In vector form, the right hand side is the vector function $f=(y_2,(57+\sin t)y_1)$. The Jacobian of this function is $$J(f)=\begin{pmatrix}0 &1\\(57+\sin t) &0 \end{pmatrix}$$
The eigenvalues are $\lambda=+-\sqrt{(57+\sin t)}$. So the eigenvalues are not all negative, and their ratio is $-1$. Is this enough to deduce that this problem is stiff? Investigating the behaviour of numerical solutions is not straightforward, because there are no initial values.
 A: As far as I am aware there isn't really a precise definition for whether an equation is stiff or not, and so the main way of determining this would be to choose an arbitrary initial condition, model it numerically then repeat this for a small change in the initial conditions. If the solution varies wildly, then this can be considered stiff. However, I would say that the presence of the sine in there would have an effect on its stiffness.
As you determined you have the system:
$$\frac{d}{dt}\mathbf{y}=\mathbf{M}\mathbf{y}$$
so in your case try various values for $y_1,y_2$ at $t=0$ as this will be the easiest to evaluate.
A: Stiffness is a combined property of ODE, numerical method and numerical number type, the latter could be replaced by the admissible range of step sizes. It occurs when the step size selection is dominantly restricted by the stability region of the method, and less by the desired error target.
For this problem it would seem that step size restrictions are more related to the rapid exponential growth of the solution than to the stability region of the method.
