1 Answer

Colen · Answer 1 · 2020-12-02T06:49:13+0000

1. Filling in the table is just a matter of plugging in the given
x,y values into the ODE
$(\mathrm dy)/(\mathrm dx)=\frac{xy}3$ :

$\begin{array}cx&-1&-1&-1&0&0&0&1&1&1\\ y&1&2&3&1&2&3&1&2&3\\(\mathrm dy)/(\mathrm dx)&-\frac13&-\frac23&-1&0&0&0&\frac13&\frac23&1\end{array}$

2. I've attached what the slope field should look like. Basically, sketch a line of slope equal to the value of
$(\mathrm dy)/(\mathrm dx)$ at the labeled point (these values are listed in the table).

3. This ODE is separable. We have

$(\mathrm dy)/(\mathrm dx)=\frac{xy}3\implies\frac{\mathrm dy}y=\frac x3\,\mathrm dx$

Integrating both sides gives

$\ln|y|=\frac{x^2}6+C\implies y=e^(x^2/6+C)=Ce^(x^2/6)$

With the initial condition
f(0)=4 , we take
x=0 and
y=4 to solve for
:

$4=Ce^0\implies C=4$

Then the particular solution is

$\boxed{y=4e^(x^2/6)}$

4. First, solve the ODE (also separable):

$(\mathrm dT)/(\mathrm dt)=k(T-38)\implies(\mathrm dT)/(T-38)=k\,\mathrm dt$

Integrating both sides gives

$\ln|T-38|=kt+C\implies T=38+Ce^(kt)$

Given that
T(0)=75 , we can solve for
:

$75=38+C\implies C=37$

Then use the other condition,
T(30)=60 , to solve for
:

$60=38+37e^(30k)\implies k=\frac1{30}\ln(22)/(37)$

Then the particular solution is

$T(t)=38+37e^{\left(\frac1{30}\ln(22)/(37)\right)t}$

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

$T(60)=38+37e^{2\ln(22)/(37)}\approx\boxed{51^\circ}$