Question

obtain the differential equation of the family of curve and sketch several representative members of the family:

Parabolas with axis parallel to x-axis

Answer 1

my teacher already give the answer.. and this is the answer

Answer 2

your teacher got
2y'* y'' - 3 y ''' = 0

Answer 3

can you take a screenshot of that whole sheet

Answer 4

none

Answer 5

left side no.2

Answer 6

since it says the parabola is parallel to the x axis
(sorry my earlier work is incorrect)

Answer 7

it's okay..
no need to apologize

Answer 8

actually to be more general, since it says just parallel to the x axis
(y-k)^2 = 4a (x-b)

Answer 9

now we have three arbitrary constants, which we need to eliminate

Answer 10

okay, the n next..

Answer 11

(y-k)^2 = 4a (x-b)

2(y-k) * y ' = 4a * 1

2(y' ) * y' + 2(y-k) y' ' = 0

Answer 12

y^2 - 2ky + k^2 = 4ax-4ab

where k^2 and -4ab will become zero..

Answer 13

am i right?

Answer 14

yes,

Answer 15

(y-k)^2 = 4a (x-b)

y^2 - 2ky + k^2 = 4ax-4ab

2y y' - 2k y ' + 0 = 4a

2y' y' + 2y y' ' - 2k y ' = 0

Answer 16

we still have that k, so we have to take derivative again

Answer 17

yes..

Answer 18

isn't it -ky''??

Answer 19

-2ky''?

Answer 20

yes

Answer 21

(y-k)^2 = 4a (x-b)

y^2 - 2ky + k^2 = 4ax-4ab

2y y' - 2k y ' + 0 = 4a

2y' y' + 2y y' ' - 2k y '' = 0

divide through by 2 , simplify

(y ' )^2 + y * y ' ' - k y ' ' = 0

Answer 22

transpose -ky'' to the right side then divide by y''??

Answer 23

thats a great idea

Answer 24

what's next??

Answer 25

where??

Answer 26

$$\Large{ (y ' )^2 + y \cdot y ' ' = k y ' ' \\

(y ' )^2 + y \cdot y ' ' = k y ' '\\

\frac{(y ' )^2}{y''} + \frac{y \cdot y ' '}{y''} = k \\

\frac{(y ' )^2}{y''} + y = k \\~\\

\frac{y ''~ 2(y ' )\color{red}{y ''} - (y')^2 y'''}{(y'')^2} + y' = 0 \\

}$$

Answer 27

awhh, okay..

Answer 28

then we can find a common denominator

Answer 29

$$\Large{ (y ' )^2 + y \cdot y ' ' = k y ' ' \\

(y ' )^2 + y \cdot y ' ' = k y ' '\\

\frac{(y ' )^2}{y''} + \frac{y \cdot y ' '}{y''} = k \\

\frac{(y ' )^2}{y''} + y = k \\~\\

\frac{y ''~ 2(y ' )\color{red}{y ''} - (y')^2 y'''}{(y'')^2} + y' = 0 \\

\frac{y ''~ 2(y ' )\color{red}{y ''} - (y')^2 y'''}{(y'')^2} + y' \cdot \frac{(y'')^2}{(y'')^2}= 0 \\

}$$

Answer 30

multiply both sides by ( y ' ' ) ^2

Answer 31

in order to cancel the y''??

Answer 32

in denominator??

Answer 33

$$\Large{ (y ' )^2 + y \cdot y ' ' = k y ' ' \\

(y ' )^2 + y \cdot y ' ' = k y ' '\\

\frac{(y ' )^2}{y''} + \frac{y \cdot y ' '}{y''} = k \\

\frac{(y ' )^2}{y''} + y = k \\~\\

\frac{y ''~ 2(y ' ){y ''} - (y')^2 y'''}{(y'')^2} + y' = 0 \\

\frac{y ''~ 2(y ' ){y ''} - (y')^2 y''' + y' \cdot (y'')^2}{(y'')^2 } = 0 \\~\\

y ''~ 2(y ' ){y ''} - (y')^2 y''' + y' \cdot (y'')^2=0\\~\\
y ''~ 2(y ' ){y ''} - (y')^2 y''' + y' \cdot (y'')^2=0\\~\\
2y' (y ' ' )^2 - (y')^2 y ''' +y' \cdot (y'')^2=0\\~\\
2y' (y ' ' )^2 +y' \cdot (y'')^2- (y')^2 y ''' = 0 \\~\\
3y' (y ' ' )^2 - (y')^2 y ''' = 0 \\~\\


}$$

Answer 34

can you read what your teacher wrote

Answer 35

yes, y'(y''')-3(y'')^2=0

Answer 36

we almost got that answer

Answer 37

$$
\Large{ (y ' )^2 + y \cdot y ' ' = k y ' ' \\

(y ' )^2 + y \cdot y ' ' = k y ' '\\

\frac{(y ' )^2}{y''} + \frac{y \cdot y ' '}{y''} = k \\

\frac{(y ' )^2}{y''} + y = k \\~\\

\frac{y ''~ 2(y ' ){y ''} - (y')^2 y'''}{(y'')^2} + y' = 0 \\

\frac{y ''~ 2(y ' ){y ''} - (y')^2 y''' + y' \cdot (y'')^2}{(y'')^2 } = 0 \\~\\

y ''~ 2(y ' ){y ''} - (y')^2 y''' + y' \cdot (y'')^2=0\\~\\
y ''~ 2(y ' ){y ''} - (y')^2 y''' + y' \cdot (y'')^2=0\\~\\
2y' (y ' ' )^2 - (y')^2 y ''' +y' \cdot (y'')^2=0\\~\\
3y' (y ' ' )^2 - (y')^2 y '''=0\\~\\



}
$$

Answer 38

oh divide by y '

Answer 39

almost got it.. but i still wonder why this is the answer, i already solve this but still we both have the same answer..

Answer 40

gotcha..
thank you so much..

Answer 41

$$
\Large{ (y ' )^2 + y \cdot y ' ' = k y ' ' \\

(y ' )^2 + y \cdot y ' ' = k y ' '\\

\frac{(y ' )^2}{y''} + \frac{y \cdot y ' '}{y''} = k \\

\frac{(y ' )^2}{y''} + y = k \\~\\

\frac{y ''~ 2(y ' ){y ''} - (y')^2 y'''}{(y'')^2} + y' = 0 \\

\frac{y ''~ 2(y ' ){y ''} - (y')^2 y''' + y' \cdot (y'')^2}{(y'')^2 } = 0 \\~\\

y ''~ 2(y ' ){y ''} - (y')^2 y''' + y' \cdot (y'')^2=0\\~\\
y ''~ 2(y ' ){y ''} - (y')^2 y''' + y' \cdot (y'')^2=0\\~\\
2y' (y ' ' )^2 - (y')^2 y ''' +y' \cdot (y'')^2=0\\~\\
3y' (y ' ' )^2 - (y')^2 y '''=0\\~\\
\frac{3y' (y ' ' )^2 - (y')^2 y '''}{y'}=\frac{0}{y'}\\~\\
3(y ' ' )^2 - y' \cdot y '''=0 \\ ~\\\rm divide ~ by~ sides~ by -1\\~\\
y' \cdot y '''-3(y ' ' )^2=0


}
$$

Answer 42

me too, i wonder why...

Answer 43

thank youu so much.,

Answer 44

and to graph the family of curves you know how to do that .