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2 years ago

11

A car covers a distance of 231.36 km in 3.2 hrs. Find the distance covered by the car in 1 hour. Also, how much distance will it

cover in 5 hours?

1 answer:

Anit [1.1K]2 years ago

7 0

Answer:

The distance in 1 hr is 72.3 km/hr and the distance in 5 hrs is 361.5 km

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Jayla buys and sells vintage clothing. She bought two blouses for $25.00 each and later sold them for $38.00 each. She bought th

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Solve the given initial value problem and determine how the interval in which the solution exists depends on the initial value y

zalisa [80]

Answer:

y has a finite solution for any value y_0 ≠ 0.

Step-by-step explanation:

Given the differential equation

y' + y³ = 0

We can rewrite this as

dy/dx + y³ = 0

Multiplying through by dx

dy + y³dx = 0

Divide through by y³, we have

dy/y³ + dx = 0

dy/y³ = -dx

Integrating both sides

-1/(2y²) = - x + c

Multiplying through by -1, we have

1/(2y²) = x + C (Where C = -c)

Applying the initial condition y(0) = y_0, put x = 0, and y = y_0

1/(2y_0²) = 0 + C

C = 1/(2y_0²)

So

1/(2y²) = x + 1/(2y_0²)

2y² = 1/[x + 1/(2y_0²)]

y² = 1/[2x + 1/(y_0²)]

y = 1/[2x + 1/(y_0²)]½

This is the required solution to the initial value problem.

The interval of the solution depends on the value of y_0. There are infinitely many solutions for y_0 assumes a real number.

For y_0 = 0, the solution has an expression 1/0, which makes the solution infinite.

With this, y has a finite solution for any value y_0 ≠ 0.

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2 years ago

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Answer:

Step-by-step explanation:

1 ) Given that

$(d^2y/dx^2) + 4y = x - x^2 + 20\\\\ (d^2y/dx^2) + 4y = - x^2 + x + 20$

For a non homogeneous part $- x^2 + x + 20$ , we assume the particular solution is

$y_p(x) = Ax^2 + Bx + C$

2 ) Given that

$d^2y/dx^2 + 6dy/dx + 8y = e^{2x}$

For a non homogeneous part $e^{2x}$ , we assume the particular solution is

$y_p(x) = Ae^{2x}$

3 ) Given that

y′′ + 4y′ + 20y = −3sin(2x)

For a non homogeneous part −3sin(2x) , we assume the particular solution is

$y_p(x) = Acos(2x)+Bsin(2x)$

4 ) Given that

y′′ − 2y′ − 15y = 3xcos(2x)

For a non homogeneous part 3xcos(2x) , we assume the particular solution is

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1 year ago