For the following initial value problem, compute the first two approximations $u_1$ and $u_2$ given by Euler's method using the given time step.
$$y^{\prime }(t)=-y,y(0)=4;\mathrm{\Delta }t=0.3$$
$u_1=$
(Simplify your answer.)
$$u_2=$$
(Simplify your answer.)

Step 1 ：We are given the initial value problem $y^{\prime }(t)=-y$ with $y(0)=4$ and $\mathrm{\Delta }t=0.3$. We are asked to compute the first two approximations, $u_1$ and $u_2$, using Euler's method.

Step 2 ：Euler's method is a numerical method used to approximate solutions to first order ordinary differential equations. It is based on the idea of using tangent lines to approximate the curve of the solution. The formula for Euler's method is $u_{n+1}=u_n+\mathrm{\Delta }t\cdot f(t_n,u_n)$, where $f(t_n,u_n)$ is the derivative of the function at the point $(t_n,u_n)$, and $\mathrm{\Delta }t$ is the step size.

Step 3 ：In this case, the derivative $f(t,y)=-y$, the initial value $y(0)=4$, and the step size $\mathrm{\Delta }t=0.3$.

Step 4 ：To find $u_1$ and $u_2$, we can use the formula above: $u_1=u_0+\mathrm{\Delta }t\cdot f(t_0,u_0)=4+0.3\cdot -4$ and $u_2=u_1+\mathrm{\Delta }t\cdot f(t_1,u_1)=u_1+0.3\cdot -u_1$.

Step 5 ：Calculating these values, we find that $u_1=2.8$ and $u_2=1.96$.

Step 6 ：Final Answer: The first two approximations given by Euler's method are $u_1=\boxed{{\displaystyle 2.8}}$ and $u_2=\boxed{{\displaystyle 1.96}}$.