Please explain and show work

Mathematics

2 answers:

zaharov [31]4 years ago

7 0

Let's solve your equation step-by-step.(1 3/5)(t−6)= −0.4Step 1: Simplify both sides of the equation.8/5t + −48/5= −0.4Step 2: Add 48/5 to both sides.8/5t + −48/5 + 48/5 = −0.4 + 48/58/5t = 46/5Step 3: Multiply both sides by 5/8.(5/8)*(8/5t)=(5/8)*(46/5)t=23/4Answer:t = 5 3/4

Vinvika [58]4 years ago

6 0

1*3/5 = 3/5.
3/5(t-6) = -0.4
Distribute 3/5
3/5t - 3.6 = -0.4
Add 3.6
3/5t = 3.2
Divide by 3/5
t = 5 1/3 or 5.33

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Consider the following equation of the form dy/dt = f(y)dy/dt = ey − 1, −[infinity] < y0 < [infinity](a) Sketch the graph

kotegsom [21]

Complete Question:

The complete question is shown on the first uploaded image

Answer:

a) The graph of f(y) versus y. is shown on the second uploaded image

b) The critical point is at y = 0 and the solution is asymptotically unstable.

c)The phase line is shown on the third uploaded image

d) The sketch for the several graphs of solution in the ty-plane is shown on the fourth uploaded image

Step-by-step explanation:

Step One: Sketch The Graph of f(y) versus y

Looking at the given differential equation

$\frac{dy}{dt} = e^{y} - 1$ for -∞ < $y_{o}$ < ∞

We can say let $\frac{dy}{dt}$ = f(y) = $e^{y} - 1$

Now the dependent value is f(y) and the independent value is y so to sketch is graph we can assume a scale in this case i cm on the graph is equal to 2 unit for both f(y) and y and the match the coordinates and after that join the point to form the graph as shown on the uploaded image.

Step Two : Determine the critical point

To fin the critical point we have to set $\frac{dy}{dt}$ = 0

This means $e^{y} - 1$ = 0

For this to be possible $e^{y}$ = 1

which means that $e^{y}$ = $e^{0}$

which implies that y = 0

Hence the critical point occurs at y = 0

meaning that the equilibrium solution is y = 0

As t → ∞, our curve is going to move away from y = 0 hence it is asymptotically unstable.

Step Three : Draw the Phase lines

A phase line can be defined as an image that shows or represents the way an ODE(ordinary differential equation ) that does not explicitly depend on the independent variable behaves in a single variable. To draw this phase line , draw the y-axis as a vertical line and mark on it the equilibrium, i.e. where f(y) = 0.

In each of the intervals bounded by the equilibrium draw an upward

pointing arrow if f(y) > 0 and a downward pointing arrow if f(y) < 0.

This phase line would solely depend on y does not matter what t is

On the positive x axis it would get steeper very quickly as you move up (looking at the part A graph).

For below the x-axis which stable (looking at the part a graph) we are still going to have negative slope but they are going to be close to 0 and they would take a little bit longer to get steeper

Step Four : Draw a Solution Curve

A solution curve is a curve that shows the solution of a DE (deferential equation)

Here the solution curve would be drawn on the ty-plane

So the t-axis(x-axis) is its the equilibrium that is it is the solution

If we are above the x-axis it is going to increase faster and if we are below it is going to decrease but it would be slower (looking at part A graph)