midterm 1 fix three questions

TASK: MATH

INSTRUCTOR:

DATE:

1. Find the phase lines for the following differential equations and use these to sketch solutions for them in the slope field. Classify each equilibrium point as a source, sink or node.

Hence the solutions will be

To find the phase lines, we use the above solutions for y. 1

The phase diagram is therefore:

The sketch solution is:

,

The phase line diagram, is:

Sketching the solution:

Qn. 2). Suppose that the graph below is a solution to the differential equation

i). How much of the slope field can you sketch from this information? Assume that f(y) is continuous everywhere and note that the equation is autonomous. Sketch it on the graph above.

Solution

For autonomous equations, the slope is similar in the horizontal direction. The slope at is same to slope at for numbers . The solution is y(x) is positive and converges to zero. This is when x is approaching infinity. Mini-tangents drawn to the curve of y(x) represent the slope field for all points of y greater than zero. All slopes along x-axis are also zero. The graph could also converge to some another point such as. In this case, the slope of the field at all points with y greater than y*

ii) What can you say about a solution that has Sketch it on the graph above?

The solution shifts horizontally to the left by units.

iii) Determine if the solution of an autonomous differential equation can have a local max or min. either provide an example of this or explain why it can’t occur

The solution of an autonomous differential equation can either have a local max or min

3. For the following differential equations, find all bifurcation values, draw phase lines before, after, and at the bifurcation values, and use this to complete a bifurcation diagram

When , no equilibrium points are there. When, the equilibrium point exists at when the equilibrium points are two. These are .

Hence is a bifurcation value.

For , In this case the solutions are increasing

For , except when is a node.

When , for all

For and too.

The phase lines are:

(ii). ) (1)

Solving the above equation;

Integrating;

;

Solution of value of y will be given by quadratic equation:

:

If

4. Solve the following initial value problems using the technique of integrating factors. Graph the slope field for the equation and plot your answer on it, using Desmos to assist you.

i) y (0)=1

To find the integrating factor,

Hence integrating factor I=

Therefore

Integrating both sides,

But , Therefore C=

And

Hence

ii) y (0)=

Integrating both sides;

Therefore C=

5. Suppose a 100 litre tank is full of saltwater. Two tubes are dispensing saltwater into the tank. Tube one has salt water at a concentration of 2kg per litre, and its flowing at 6 litres per minute. Tube two has variable concentration of kg of salt per litre, and it’s flowing at 4 litres per minute.

i) Assume well mixed water is flowing out of the tank to keep it full. Set up a differential equation describing this scenario.

C: Concentration of salt in the tank () and ()

t: time (min)

Rate of change of concentration with time, =rate in –rate out

Rate in=positive change in concentration

Rate of salt entering tank=)+) =

Rate in= But

Rate out=C =5C.

ii. =1.5 what will the amount of salt in the tank tend to over time?

Rate of salt entering tank= But

Amount of salt entering tank=12+6=

iii. If S(0)=160 and =0.5,will the amount of salt increase or decrease over time? Why?

If =0.5, -5C

-800=-799.86Hence the amount will decrease because rate out is more than rate in

5. Assume now that the tank starts half full, so there is 50 liters of liquid in the tank at t=0. Suppose that well-mixed water is flowing out of the tank at a rate of 8 liters per minute, and that the tube 2 are the same as before, with variable.

iv) Revise your differential equation from I to include this information

At t=0, 50 Litres

Rate out=8 liters/min, Amount leaving=2C

C: Concentration of salt in the tank () and ()

t: time (min)

Rate of change of concentration with time, =rate in –rate out

.

But 2C

v) Solve your equation using the method of integrating factors. Then find the specific solution where S(0)=0 . If =2, approximately how much salt is in the tank when it is full?

If S(0)=0 . If =2

Using Integrating factors: The integral is =

But S(0)=0 Therefore C=0.

Hence C=0.15

Y-Values -2 0 5 7 7 3 0.3 0.1

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Y-Values -2 0 5 7 7 3 0.3 0.1 R -2 0 5 7 6 2 0.1

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Y-Values -2 0 1.5 2.5 -2 3 -2 3

Y-Values 0 3 4 0 3 -4 -1 0 0 2

Y-Values 0 3 4 3 0 0.5 3 -1.4 -0.1 0 0.1 1.4 1.5 2