Answers

$$\begin{aligned} \mathbf{1}(\mathbf{u}(\mathbf{x},t) = t + e^{-\mathbf{x}}, \ \ \ \ \mathbf{2})\mathbf{u}(\mathbf{x},t) &= t^2 + \mathbf{x}t, \ \ \ \mathbf{3}\ \ \mathbf{u}(\mathbf{x},t) = 1 + \mathbf{x}^2t^2, \ \ \ \mathbf{4})\mathbf{u}(\mathbf{x},t) = t + \sin\mathbf{x}t \\ \ \ \ \mathbf{5})\mathbf{u}(\mathbf{x},t) &= \frac{\mathbf{x}}{t-1}, \ \ \ \mathbf{6})\mathbf{u}(\mathbf{x},t) = \mathbf{x} + e^t, \ \ \ \ \mathbf{7})\mathbf{u}(\mathbf{x},t) = \mathbf{1} + \mathbf{x}t, \ \ \ \mathbf{8})\mathbf{u}(\mathbf{x},t) = \mathbf{1} + \mathbf{x}t \\ \ \ \ \ \ \ \ \ \ \ \ \end{aligned}$$

$$\begin{aligned} \ \ \ \ \ \ \ \ \ \end{aligned} \end{aligned} \begin{aligned} \ \ \ \ \ \ \ \ \ \ \end{aligned} \begin{aligned} \ \ \ \ \ \ \ \ \end{aligned} \begin{aligned} \ \ \ \ \ \ \ \ \end{aligned} \begin{aligned} \ \ \ \ \ \ \ \end{aligned} \begin{aligned} \ \ \ \ \ \ \ \end{aligned} \begin{aligned} \ \ \ \ \ \ \ \end{aligned} \begin{aligned} \ \ \ \ \ \ \ \ \ \end{aligned} \begin{aligned} \ \ \ \ \ \ \ \ \ \ \ \end{aligned}$$
