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Yuri [45]
3 years ago
10

Consider the following differential equation. x^2y' + xy = 3 (a) Show that every member of the family of functions y = (3ln(x) +

C)/x is a solution of the differential equation. (Do this on paper. Your instructor may ask you to turn in this work.) (b) Illustrate part (a) by graphing several members of the family of solutions on a common screen. (Do this on paper. Your instructor may ask you to turn in this work.) (c) Find a solution of the differential equation that satisfies the initial condition y(1) = 3. (Enter the argument of the logarithmic function in parentheses.) y(x) = (d) Find a solution of the differential equation that satisfies the initial condition y(3) = 1. (Enter the argument of the logarithmic function in parentheses.) y(x) =
Mathematics
1 answer:
Veronika [31]3 years ago
6 0

Answer:

Verified

y(x) = \frac{3Ln(x) + 3}{x}

y(x) = \frac{3Ln(x) + 3 - 3Ln(3)}{x}

Step-by-step explanation:

Question:-

- We are given the following non-homogeneous ODE as follows:

                           x^2y' +xy = 3

- A general solution to the above ODE is also given as:

                          y = \frac{3Ln(x) + C  }{x}

- We are to prove that every member of the family of curves defined by the above given function ( y ) is indeed a solution to the given ODE.

Solution:-

- To determine the validity of the solution we will first compute the first derivative of the given function ( y ) as follows. Apply the quotient rule.

                          y' = \frac{\frac{d}{dx}( 3Ln(x) + C ) . x - ( 3Ln(x) + C ) . \frac{d}{dx} (x)  }{x^2} \\\\y' = \frac{\frac{3}{x}.x - ( 3Ln(x) + C ).(1)}{x^2} \\\\y' = - \frac{3Ln(x) + C - 3}{x^2}

- Now we will plug in the evaluated first derivative ( y' ) and function ( y ) into the given ODE and prove that right hand side is equal to the left hand side of the equality as follows:

                          -\frac{3Ln(x) + C - 3}{x^2}.x^2 + \frac{3Ln(x) + C}{x}.x = 3\\\\-3Ln(x) - C + 3 + 3Ln(x) + C= 3\\\\3 = 3

- The equality holds true for all values of " C "; hence, the function ( y ) is the general solution to the given ODE.

- To determine the complete solution subjected to the initial conditions y (1) = 3. We would need the evaluate the value of constant ( C ) such that the solution ( y ) is satisfied as follows:

                         y( 1 ) = \frac{3Ln(1) + C }{1} = 3\\\\0 + C = 3, C = 3

- Therefore, the complete solution to the given ODE can be expressed as:

                        y ( x ) = \frac{3Ln(x) + 3 }{x}

- To determine the complete solution subjected to the initial conditions y (3) = 1. We would need the evaluate the value of constant ( C ) such that the solution ( y ) is satisfied as follows:

                         y(3) = \frac{3Ln(3) + C}{3} = 1\\\\y(3) = 3Ln(3) + C = 3\\\\C = 3 - 3Ln(3)

- Therefore, the complete solution to the given ODE can be expressed as:

                        y(x) = \frac{3Ln(x) + 3 - 3Ln(3)}{y}

                           

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Draw a rectangular fraction model to explain your thinking. Then, write a multiplication sentence. A. 2 3 of = 3 5
myrzilka [38]

Answer:

\frac{2}{3}\ of\ \frac{3}{5} = \frac{2}{5}

Step-by-step explanation:

Given

\frac{2}{3}\ of\ \frac{3}{5}

Solving (a): Multiplication sentence

\frac{2}{3}\ of\ \frac{3}{5}

Rewrite of as *

\frac{2}{3}\ of\ \frac{3}{5} = \frac{2}{3}\ *\ \frac{3}{5}

Solve

\frac{2}{3}\ of\ \frac{3}{5} = \frac{2}{5}

Hence, the multiplication sentence is:

\frac{2}{3}\ of\ \frac{3}{5} = \frac{2}{5}

Solving (b): Rectangular fraction model

In (a) above, the result is 2/5

The fraction model will be represented as thus:

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<em>See attachment for model</em>

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3 years ago
RSM Question, please help!
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mojhsa [17]

Answer:

a) The equation is (y - 1)² = -8 (x - 4)

b) The equation is (x - 1)²/25 + (y - 4)²/16 = 1

c) The equation of the ellipse is (x - 3)²/16 + y²/4 = 1

Step-by-step explanation:

a) Lets revise the standard form of the equation of the parabola with a

   horizontal axis

# (y - k)² = 4p (x - h), (h , k) are the coordinates of its vertex and p ≠ 0

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* Lets solve the problem

∵ The focus is (2 , 1)

∵ focus is (h + p , k)

∴ h + p = 2 ⇒ subtract p from both sides

∴ h = 2 - p ⇒ (1)

∴ k = 1

∵ It opens left, then the axis is horizontal and p is negative

∴ Its equation is (y - k)² = 4p (x - h)

∵ k = 1

∴ Its equation is (y - 1)² = 4p (x - h)

- The parabola contains point (2 , 5), substitute the coordinates of the

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∴ (5 - 1)² = 4p (2 - h)

∴ (4)² = 4p (2 - h)

∴ 16 = 4p (2 - h) ⇒ divide both sides by 4

∴ 4 = p (2 - h) ⇒ (2)

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∴ 4 = p (2 - [2 - p]) ⇒ open the inside bracket

∴ 4 = p (2 - 2 + p) ⇒ simplify

∴ 4 = p (p)

∴ 4 = p² ⇒ take √ for both sides

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∵ h = 2 - p

∵ p = -2

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∴ The equation of the parabola in standard form is

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∴ The equation is (y - 1)² = -8 (x - 4)

b) Lets revise the equation of the ellipse

- The standard form of the equation of an ellipse with  center (h , k)

 and major axis parallel to x-axis is (x - h)²/a² + (y - k)²/b² = 1  

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- The coordinates of the foci are (h ± c , k), where c² = a² - b²  

* Now lets solve the problem

∵ Its vertices are (-4 , 4) and (6 , 4)

∵ The coordinates of the vertices are (h + a , k ) and (h - a , k)  

∴ k = 4

∴ h + a = 6 ⇒ (1)

∴ h - a = -4 ⇒ (2)

- Add (1) and (2) to find h

∴ 2h = 2 ⇒ divide both sides by 2

∴ h = 1

- Substitute the value of h in (1) or (2) to find a

∴ 1 + a = 6 ⇒subtract 1 from both sides

∴ a = 5

∵ The foci at (-2 , 4) and (4 , 4)

∵ The coordinates of the foci are (h + c , k) , (h - c , k)

∴ h + c = 4

∵ h = 1

∴ 1 + c = 4 ⇒ subtract 1 from both sides

∴ c = 3

∵ c² = a² - b²

∴ 3² = 5² - b²

∴ 9 = 25 - b² ⇒ subtract 25 from both sides

∴ -16 = -b² ⇒ multiply both sides by -1

∴ 16 = b²

∵ a² = 25

∵ The equation of the ellipse is (x - h)²/a² + (y - k)²/b² = 1

∴ The equation is (x - 1)²/25 + (y - 4)²/16 = 1

c) How to identify the type of the conic  

- Rewrite the equation in the general form,  

 Ax² + Bxy + Cy² + Dx + Ey + F = 0  

- Identify the values of A and C from the general form.  

- If A and C are nonzero, have the same sign, and are not equal  

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- If A and C are equal and nonzero and have the same sign, then

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* Now lets solve the problem

∵ x² + 4y² - 6x - 7 = 0

∵ The general form of the conic equation is

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∵ If A and C are nonzero, have the same sign, and are not equal  to

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∵ x² + 4y² - 6x - 7 = 0 ⇒ re-arrange the terms

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∴ (x² - 6x + 9) - 9 + 4y² - 7 = 0 ⇒ simplify

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∴ (x - 3)² + 4y² = 16 ⇒ divide all terms by 16

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