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lyudmila [28]
2 years ago
13

1. (a) Solve the differential equation (x + 1)Dy/dx= xy, = given that y = 2 when x = 0. (b) Find the area between the two curves

y = x2 and y = 2x – x2.​

Mathematics
1 answer:
erastova [34]2 years ago
7 0

(a) The differential equation is separable, so we separate the variables and integrate:

(x+1)\dfrac{dy}{dx} = xy \implies \dfrac{dy}y = \dfrac x{x+1} \, dx = \left(1-\dfrac1{x+1}\right) \, dx

\displaystyle \frac{dy}y = \int \left(1-\frac1{x+1}\right) \, dx

\ln|y| = x - \ln|x+1| + C

When x = 0, we have y = 2, so we solve for the constant C :

\ln|2| = 0 - \ln|0 + 1| + C \implies C = \ln(2)

Then the particular solution to the DE is

\ln|y| = x - \ln|x+1| + \ln(2)

We can go on to solve explicitly for y in terms of x :

e^{\ln|y|} = e^{x - \ln|x+1| + \ln(2)} \implies \boxed{y = \dfrac{2e^x}{x+1}}

(b) The curves y = x² and y = 2x - x² intersect for

x^2 = 2x - x^2 \implies 2x^2 - 2x = 2x (x - 1) = 0 \implies x = 0 \text{ or } x = 1

and the bounded region is the set

\left\{(x,y) ~:~ 0 \le x \le 1 \text{ and } x^2 \le y \le 2x - x^2\right\}

The area of this region is

\displaystyle \int_0^1 ((2x-x^2)-x^2) \, dx = 2 \int_0^1 (x-x^2) \, dx = 2 \left(\frac{x^2}2 - \frac{x^3}3\right)\bigg|_0^1 = 2\left(\frac12 - \frac13\right) = \boxed{\frac13}

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Sarawongs school is selling tickets to the anual talent show on the first day of ticket sales the school sold 1 adult ticket and
vaieri [72.5K]

Price of one adult ticket is $9 and price of one student ticket is also $9.

Step-by-step explanation:

Let,

Adult ticket = x

Student ticket = y

According to given statement;

x+2y=27   Eqn 1

2x+2y=36   Eqn 2

Subtracting Eqn 1 from Eqn 2;

(2x+2y)-(x+2y)=36-27\\2x+2y-x-2y=9\\x=9

Putting x=9 in Eqn 1;

9+2y=27\\2y=27-9\\2y=18

Dividing both sides by 2;

\frac{2y}{2}=\frac{18}{2}\\y=9

Price of one adult ticket is $9 and price of one student ticket is also $9.

Keywords: linear equations, subtraction

Learn more about linear equations at:

  • brainly.com/question/10081622
  • brainly.com/question/10341324

#LearnwithBrainly

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Eden just bought a trough in the shape of a rectangular prism for her horses. She needs to know what volume of water to add to t
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The volume of the trough is V(w) = w³ + 20w² - 429w and the rate of change of the volume over a width of 38 inches to 53 inches is 4695 in³/in

<h3>What is an equation?</h3>

An equation is an expression that shows the relationship between two or more variables and numbers.

Let w represent the width, hence:

length = w + 33, height = w - 13

Volume (V) = w(w + 33)(w - 13) = w³ + 20w² - 429w

V(w) = w³ + 20w² - 429w

Rate of change = dV/dw = 3w² + 40w - 429

When w = 38, dV/dw = 3(38)² + 40(38) - 429 = 5423

When w = 53, dV/dw = 3(53)² + 40(53) - 429 = 10118

Rate = 10118 - 5423 = 4695 in³/in

The volume of the trough is V(w) = w³ + 20w² - 429w and the rate of change of the volume over a width of 38 inches to 53 inches is 4695 in³/in

Find out more on equation at: brainly.com/question/2972832

#SPJ1

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