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yuradex [85]
3 years ago
7

I neeed help!!!!!!!!!!!!!

Mathematics
2 answers:
Illusion [34]3 years ago
7 0
I think its b -3/10 • -6
ikadub [295]3 years ago
7 0
Answer is c the poin means ( + )
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Consider the following differential equation. x^2y' + xy = 3 (a) Show that every member of the family of functions y = (3ln(x) +
Veronika [31]

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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6 0
3 years ago
What is the value of x drawing not to scale
AVprozaik [17]
 angle x would end up being equal to 71 degrees.
8 0
3 years ago
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In a fifth grade class, 4/5 of the girls have brown hair, of the brown-haired girls, 3/4
AfilCa [17]

Answer:

\frac{3}{5}

Step-by-step explanation:

In the fifth grade class, \frac{4}{5} of the girls have brown hair. Again among those of the brown-haired girls, \frac{3}{4} of them have long hair.

We are asked to find the fraction of the girls who have long brown hair in the class.

The fraction of the total students which have long brown hair will be (\frac{4}{5} \times \frac{3}{4}) = \frac{3}{5} . (Answer)

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3 years ago
Sweets cost 6p how much would 43 sweets cost?
SCORPION-xisa [38]
The answer is 258 pennies
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A. Are the expressions 3(m - 2) + 2(m - 2) and 5(m - 2) equivalent expressions?
Lena [83]

Answer:

Step A. 3(m - 2) + 2(m - 2) and 5(m - 2) are equivalent expressions.

Step-by-step explanation:

Step B. If you calculate 3(m - 2) + 2(m - 2) you will see it will equal to 5(m - 2) therefore, they are equal.

Hope I Helped I'm New! :D

8 0
3 years ago
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