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love history [14]
3 years ago
13

A function y(t) satisfies the differential equation dy dt = y 4 − 6y 3 + 5y 2 . (a) What are the constant solutions of the equat

ion? (Recall that these have the form y = C for some constant, C.) (b) For what values of y is y increasing? (c) For what values of y is y decreasing?
Mathematics
2 answers:
Elza [17]3 years ago
8 0

Answer:

a) y = 0 , 5,1

b) y ⊂ (- ∞,0) ∪ (0,1)∪(5,∞)

Step-by-step explanation:

Given data:

differential equation is given as

\frac{dy}[dt} = y^4 -6y^3+ 5y^2

a) constant solution

y^4 -6y^3+ 5y^2 = 0

taking y^2 from all part

y^2(y^2 - 6y -5) = 0

solution of above equation is

y = 0 , 5,1

b) for which value y is increasing

\frac{dy}{dt}  > 0

y^2(y - 5) (y -1) > 0

y ⊂ (- ∞,0) ∪ (0,1)∪(5,∞)

Alexxandr [17]3 years ago
5 0

Answer:

Hence increasing in  (-\infty,0) U (1,5)

c) Decreasing in (0,1)

Step-by-step explanation:

Given that y(t) satisfies the differential equation

\frac{dy}{dt} =y^4-6y^2+5y^2\\=y^2(y^2-6y+5)\\=y^2(y-1)(y-5)

Separate the variables to have

\frac{dy}{y^2(y-1)(y-5)} =dt

Left side we can resolve into partial fractions

Let \frac{1}{y^2(y-1)(y-5)} =\frac{A}{y} +\frac{B}{y^2}+\frac{C}{y-1} \frac{D}{y-5}

Taking LCD we get

1= Ay(y-1)(Y-5) +B(y-1)(y-5)+Cy^2 (y-5)+Dy^2 (y-1)\\Put y =1\\1 =  -4C\\Put y =5\\ 1 = 25(4)D\\Put y =0\\1=5B\\

By equating coeff of y^3 we have

A+C+D=0

C=\frac{-1}{4} \\D=\frac{1}{100} \\B =\frac{1}{5} \\A = -C-D = \frac{6}{25}

Hence left side =

\frac{6}{25y} +\frac{1}{5y^2}+\frac{-1}{4(y-1)}+ \frac{1}{100(y-5)}=dt\\\frac{6}{25}ln y -\frac{1}{5y}-\frac{1}{4}ln|(y-1)| +\frac{1}{100}ln|y-5| = t+C

b) y is increasing whenever dy/dt>0

dy/dt =0 at points y =0, 1 and 5

dy/dt >0 in (-\infty,0) U (1,5)

Hence increasing in  (-\infty,0) U (1,5)

c) Decreasing in (0,1)

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Please include answer and explanation :))
Fiesta28 [93]

Answer:

<em>Domain={2,5,8}</em>

<em>Range={1,3,6}</em>

<em>Function: YES</em>

Step-by-step explanation:

<u>The Domain of a Function </u>

Is the set of values the input variable x takes.

<u>Range of a Function </u>

It's the set of values the output function takes when x moves into the function's domain.

A function is a relation between an input and an output set of values with the condition that every element in the domain relates only with one element in the range.

Following the definitions above, the domain of the relationship is the set of input values:

Domain={2,5,8}

The range is the set of output values:

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Since each element in the input set is related only to one element in the range, the given relationship is a function.

Answers:

Domain={2,5,8}

Range={1,3,6}

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Read 2 more answers
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