Professor: For part c of #17, I believe I have found C for y(0) = 2.

However, I am not certain how to solve for y when t = 5. I have tried putting it into Sage, but the curve it has generated doesn’t seem accurate because it does not approach 4 as it should. I am not sure what Sage is doing wrong, and I cannot finish part c or d if I cannot solve for y.

My apologies for bothering you on this question again. Any help you can provide will be much appreciated.

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We solved in class (or at least highlighted the steps) and we found the implicit solution.

Use the initial condition to find c.

I don’t remember the exact numbers, but say the implicit solution is something like

1/12*ln|1-y|*(-1)+4/3*ln|1-y/4|*(-4)+ln|y|=-.25*t+c and suppose that you found c=0.3

For part (c), you plug in t=5 on the right-hand side of the equation and you get:

1/12*ln|1-y|*(-1)+4/3*ln|1-y/4|*(-4)+ln|y|=-.25*5+0.3.

You have to find y.

One way is to graph the left-hand side which is a function of y:

1/12*ln|1-y|*(-1)+4/3*ln|1-y/4|*(-4)+ln|y|

Also graph the constant function -.25*5+0.3 which appears on the right-hand side.

Plot these two graphs together and observe where they intersect. Zoom-in until you have enough precision. That is the y-value you are looking for.

For part (d), we said that you look what happens to the solution with initial condition y(0)=2 and the solution with initial condition y(0)=6. Every other solution is between these two.

I will focus on the y(0)=6 case. The other one is similar (and easier).

1. Use y(0)=6 to find c. The steps are similar to what you did for part (c).

2. Once you have found c, plug c and y=4.05 into the equation

1/12*ln|1-y|*(-1)+4/3*ln|1-y/4|*(-4)+ln|y|=-.25*t+c.

You get

1/12*ln|1-4.05|*(-1)+4/3*ln|1-4.05/4|*(-4)+ln|4.05|=-.25*t+c.

Then solve for t.

Repeat for y(0)=2.

Does this help?

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