Find \(y\) in terms of \(x\), given that: \begin{eqnarray*} \mbox{for \(x < 0\,\)}, && \frac{\d y}{\d x} = -y \mbox{ \ \ and \ \ } y = a \mbox{ when } x = -1\;; \\ \mbox{for \(x > 0\,\)}, && \frac{\d y}{\d x} = y \mbox{ \ \ \ \ and \ \ } y = b \ \mbox{ when } x = 1\;. \end{eqnarray*} Sketch a solution curve. Determine the condition on \(a\) and \(b\) for the solution curve to be continuous (that is, for there to be no `jump' in the value of \(y\)) at \(x = 0\). Solve the differential equation \[ \frac{\d y}{\d x} = \left\vert \e^x-1\right\vert y \] given that \(y=\e^{\e}\) when \(x=1\) and that \(y\) is continuous at \(x=0\,\). Write down the following limits: \ \[ \text{(i)} \ \ \lim_ {x \to +\infty} y\exp(-\e^x)\;; \ \ \ \ \ \ \ \ \ \text{(ii)} \ \ \lim_{x \to -\infty}y \e^{-x}\,. \]