Question

dy/dx=xy^3

Answer 1

Post complete Question

Answer 2

Separate variables and find the general eequation y = f(x)

Answer 3

\[\frac{\mathrm dy}{\mathrm dx}=xy^3\]
separating the varibles
\[\frac{\mathrm dy}{y^3}=x\mathrm dx\]

now integrate
\[\int\frac{\mathrm dy}{y^3}=\int x\mathrm dx\]
(dont forget the arbitrary constant of integration )

Answer 4

how do i integrate 1/y^3? :\

Answer 5

is it 1/y^2?

Answer 6

maybe it looks easier as \(y^{-3}\)

Answer 7

is it -1/y^2 or 1/y^2

Answer 8

\[\int z^n\mathrm dz=\frac{z^{n+1}}{n+1}+c\]

Answer 9

@lilfayfay we did a problem exactly like this last night....

Answer 10

Increase the power by one, divide by the new power.

Answer 11

so its -2/y^2?

Answer 12

Increase the power by one, DIVIDE by the new power

\[\Large \int\limits y^{-3}dy = \] the old power is -3, so the new power will be -2. Which you divide the term by.

Answer 13

so my answer is wrong?

Answer 14

Well.. did you divide by -2 or multiply by -2?

Answer 15

your close, try again

Answer 16

2/y^2

Answer 17

divide, by the new power, dont multiply

Answer 18

y^-2/-2?

Answer 19

Much better.

Answer 20

yeah, but you forgot the arbitrary constant of integration

Answer 21

ln|y^-2/-2|

Answer 22

?

Answer 23

am i still wrong?

Answer 24

Wait... why is it now an ln expression?
\[\Large \int\limits\limits y^{-3}dy = \frac{ y^{-2} }{ -2 } +C\]

Answer 25

arbitrary constant of integration = C, the constant. That doesn't mean add in a natural log.

Answer 26

ok but how do i integrate x then if we already added the C

Answer 27

You only need to add one C. Just integrate both sides of this, and put C on whichever side you like - it's a constant, it's not picky.\[\Large \int\limits\frac{\mathrm dy}{y^3}=\int\limits x\mathrm dx\]

Answer 28

but it says find the general equation y=f(x)

Answer 29

Yes, you'll be able to simplify to that after integrating.

Answer 30

so other side would be ln|x|

Answer 31

Why...? What is this equal to? \[\Large \int\limits \limits x dx = \]

Answer 32

1/x

Answer 33

Wait... what? Why?

Answer 34

x^2/2

Answer 35

Yes

Answer 36

y^-2/-2 + c1 = x^2/2

Answer 37

then what

Answer 38

then rearrange the equation to become y=

Answer 39

^that

Answer 40

how do i do that

Answer 41

\[\Large \frac{ y^{-2} }{ -2 } + C = \frac{ x^2 }{ 2 }\]
You need to move terms around and simplify, until you get y=

Answer 42

would y^-2/-2 become -2/y^2

Answer 43

Yes

Answer 44

nope

Answer 45

Oh, actually no. Why'd you move the -2 to the top?

Answer 46

:P

Answer 47

\[\int\frac{\mathrm dy}{y^3}=\int x\mathrm dx\]\[\frac{y^{-2}}{-2}+c =\frac{x^2}2\]
now get all the non-y terms onto the right hand side;

minus c from both sides,

then multiply both sides by -2

you should have \(y^{-2}=\)....

Answer 48

then take the reciprocal of both sides

and finally take the square root of both sides

Answer 49

show your working @lilfayfay

Answer 50

reciprocal..?

Answer 51

k, i got y^-2 = -x^2 + c1/2

Answer 52

almost, what have you done to c?

Answer 53

i multiplied both sides by -2

Answer 54

But you divided C by -2... and also, an unknown constant divided by -2 is still an unknown constant.

\[\Large C = \frac{ C }{ -2 } = 5 \times C = 10 C = C +15 = C\]
Those are all just C