1. Filling in the table is just a matter of plugging in the given [tex]x,y[/tex] values into the ODE [tex]\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{xy}3[/tex]:[tex]\begin{array}{c|ccccccccc}x&-1&-1&-1&0&0&0&1&1&1\\ y&1&2&3&1&2&3&1&2&3\\\frac{\mathrm dy}{\mathrm dx}&-\frac13&-\frac23&-1&0&0&0&\frac13&\frac23&1\end{array}[/tex]2. I've attached what the slope field should look like. Basically, sketch a line of slope equal to the value of [tex]\dfrac{\mathrm dy}{\mathrm dx}[/tex] at the labeled point (these values are listed in the table).3. This ODE is separable. We have[tex]\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{xy}3\implies\dfrac{\mathrm dy}y=\dfrac x3\,\mathrm dx[/tex]Integrating both sides gives[tex]\ln|y|=\dfrac{x^2}6+C\implies y=e^{x^2/6+C}=Ce^{x^2/6}[/tex]With the initial condition [tex]f(0)=4[/tex], we take [tex]x=0[/tex] and [tex]y=4[/tex] to solve for [tex]C[/tex]:[tex]4=Ce^0\implies C=4[/tex]Then the particular solution is[tex]\boxed{y=4e^{x^2/6}}[/tex]4. First, solve the ODE (also separable):[tex]\dfrac{\mathrm dT}{\mathrm dt}=k(T-38)\implies\dfrac{\mathrm dT}{T-38}=k\,\mathrm dt[/tex]Integrating both sides gives[tex]\ln|T-38|=kt+C\implies T=38+Ce^{kt}[/tex]Given that [tex]T(0)=75[/tex], we can solve for [tex]C[/tex]:[tex]75=38+C\implies C=37[/tex]Then use the other condition, [tex]T(30)=60[/tex], to solve for [tex]k[/tex]:[tex]60=38+37e^{30k}\implies k=\dfrac1{30}\ln\dfrac{22}{37}[/tex]Then the particular solution is[tex]T(t)=38+37e^{\left(\frac1{30}\ln\frac{22}{37}\right)t}[/tex]Now, you want to know the temperature after an additional 30 minutes, i.e. 60 minutes after having placed the lemonade in the fridge. According to the particular solution, We have[tex]T(60)=38+37e^{2\ln\frac{22}{37}}\approx\boxed{51^\circ}[/tex]5. You want to find [tex]t[/tex] such that [tex]T(t)=55[/tex]:[tex]55=38+37e^{\left(\frac1{30}\ln\frac{22}{37}\right)t}\implies\dfrac{17}{37}=e^{\left(\frac1{30}\ln\frac{22}{37}\right)t}[/tex][tex]\implies t=\dfrac{30\ln\frac{17}{37}}{\ln\frac{22}{37}}\approx\boxed{45\,\mathrm{min}}[/tex]