hilp meeee trying .... Consider the two parabolas :y1=2x^2-3x-1 and y2=x^2+7x+20. (a) Find the points of intersection of the parabolas and decide which one is greater than the other between the intersection points. (b) Compute the area enclosed by the two parabolas. (c) Use Mathcad to draw the graphs of the parabolas and find their intersection points graphically . (d) Solve the differential equation dy/dx = -exp(5-x)+3/5x^2 ; y(5)=3

(a) To find the points of intersection of the parabolas y1=2x^2-3x-1 and y2=x^2+7x+20, we need to set the two equations equal to each other and solve for x:

2x^2-3x-1 = x^2+7x+20

Rearranging the equation, we get:

x^2 + 10x + 21 = 0

To solve this quadratic equation, we can factor it as:

(x + 3)(x + 7) = 0

Setting each factor equal to zero and solving for x, we get:

x + 3 = 0 => x = -3
x + 7 = 0 => x = -7

Substituting these x-values into either of the original equations, we can find the corresponding y-values:

For y1:
When x = -3, y1 = 2(-3)^2 - 3(-3) - 1 = 24

When x = -7, y1 = 2(-7)^2 - 3(-7) - 1 = 96

For y2:
When x = -3, y2 = (-3)^2 + 7(-3) + 20 = 2

When x = -7, y2 = (-7)^2 + 7(-7) + 20 = 36

Therefore, the points of intersection are (-3, 24) and (-7, 36). To determine which one is greater, we can compare the y-values. Since 36 is greater than 24, the point of intersection (-7, 36) is greater than (-3, 24).

(b) To compute the area enclosed by the two parabolas, we need to find the integral of the difference between the two functions over the interval of intersection. Let's find the x-values where the two parabolas intersect first.

We already found that the points of intersection are (-3, 24) and (-7, 36). To find the area enclosed between the two parabolas, we need to integrate the difference of y2 - y1 over this interval.

∫[x1, x2] (y2 - y1) dx

where x1 = -3 and x2 = -7.

Substituting the equations for y1 and y2:
∫[-3, -7] (x^2+7x+20) - (2x^2-3x-1) dx

Expanding and simplifying, we get:
∫[-3, -7] -x^2 + 10x + 21 dx

Integrating term by term, we get:
[-(1/3)x^3 + 5x^2 + 21x] from -3 to -7

Evaluating this expression at -3 and -7, we get:
[-(1/3)(-3^3) + 5(-3)^2 + 21(-3)] - [-(1/3)(-7^3) + 5(-7)^2 + 21(-7)]

Simplifying further, we get:
(9/3 + 45 - 63) - (343/3 + 245 - 147) = -98/3

So, the area enclosed by the two parabolas is -98/3 square units.

(c) To graphically find the intersection points of the parabolas using Mathcad, the specific software is needed. However, you can use graphing calculators, such as Desmos or GeoGebra, to easily plot the two parabolas and determine their intersection points.

(d) To solve the differential equation dy/dx = -exp(5-x)+3/5x^2 with the initial condition y(5) = 3, we can use an integrating factor method.

Rearranging the equation, we get:

dy = (-exp(5-x) + 3/5x^2) dx

Integrating both sides, we get:

∫ dy = ∫ (-exp(5-x) + 3/5x^2) dx

y = ∫ (-exp(5-x) + 3/5x^2) dx

To solve the integral, we can split it into two parts:

y = ∫ -exp(5-x) dx + ∫ (3/5x^2) dx

Integrating each term separately:

The first term can be rewritten as:
y = -∫ exp(5-x) dx

Using the substitution method, let u = 5-x. Then du = -dx.
The integral becomes:
y = ∫ exp(u) du

Integrating, we get:
y = -exp(u) + C1

The second term can be integrated as:
y = (3/5) ∫ x^2 dx

Integrating, we get:
y = (3/5) (1/3)x^3 + C2

Combining the two terms, we have:
y = -exp(5-x) + (1/5)x^3 + C

Using the initial condition y(5) = 3, we can solve for C:
3 = -exp(5-5) + (1/5)(5^3) + C

Simplifying, we get:
3 = 1 + 5 + C
C = -3

Therefore, the solution to the differential equation is:
y = -exp(5-x) + (1/5)x^3 - 3

To solve the given problem, we will go step by step:

(a) To find the points of intersection of the parabolas, we need to solve the equation: y1 = y2.

Set up the equation: 2x^2 - 3x - 1 = x^2 + 7x + 20.

Simplify the equation: x^2 - 10x - 21 = 0.

Now, we can solve this quadratic equation either by factoring or using the quadratic formula.

Factoring: (x - 7)(x + 3) = 0.

From this, we get two possible solutions: x = 7 and x = -3.

Substitute these values back into either y1 or y2 to find the corresponding y-coordinates.

For x = 7, y1 = 2(7)^2 - 3(7) - 1 = 77 and y2 = (7)^2 + 7(7) + 20 = 77. Thus, the intersection point is (7, 77).

For x = -3, y1 = 2(-3)^2 - 3(-3) - 1 = 7 and y2 = (-3)^2 + 7(-3) + 20 = 7. Therefore, the intersection point is (-3, 7).

To determine which parabola is greater than the other between the intersection points, compare the y-values at each point.

At (7, 77), y1 = y2, so there is no clear distinction in this case.

At (-3, 7), y2 > y1. Hence, y2 is greater than y1 for x < -3.

(b) To compute the area enclosed by the two parabolas, we need to find the definite integral of their difference over the interval between the x-values of the intersection points.

Set up the integral: Area = ∫(y2 - y1) dx from -3 to 7.

Evaluate the definite integral: Area = ∫(x^2 + 7x + 20 - 2x^2 + 3x + 1) dx from -3 to 7.

Simplify the integral: Area = ∫(x^2 + 10x + 21) dx from -3 to 7.

Integrate term by term: Area = [x^3/3 + 5x^2 + 21x] from -3 to 7.

Evaluate the integral at the upper and lower limits: Area = [(7^3/3 + 5(7^2) + 21(7))] - [((-3)^3/3 + 5((-3)^2) + 21(-3))].

Calculate the result: Area = 319.

Therefore, the area enclosed by the two parabolas is 319 square units.

(c) To draw the graphs of the parabolas and find their intersection points graphically using Mathcad (or any other graphing tool):

- Open Mathcad or your preferred graphing software.
- Enter the equations y1 = 2x^2-3x-1 and y2 = x^2+7x+20.
- Plot both equations on the same graph using appropriate settings.
- The intersection points will be displayed on the graph where the two curves intersect.

(d) To solve the given differential equation dy/dx = -exp(5-x) + 3/(5x^2) with the initial condition y(5) = 3:

- Start by separating the variables to obtain: dy/(-exp(5-x) + 3/(5x^2)) = dx.
- Integrate both sides of the equation:
∫ dy/(-exp(5-x) + 3/(5x^2)) = ∫ dx.
- Solve the integral on both sides using appropriate techniques.
- Apply the initial condition y(5) = 3 to determine the constant of integration, if necessary.

Unfortunately, Mathcad is not available within this text-based platform, so I cannot provide the exact solution. However, following the steps mentioned above should lead you to the solution for the given differential equation.