Question

Solve this Separable Equation:
x(dy/dx)=1/y^3
with explanations for each step please

Answer 1

x(dy/dx) = 1/y^3 - for example

x * dy = dx / y^3 - communitive property

y^3 * x dy = dx - these aren't right

y^3 dy = dx/x - but explain each one

y^4/4 = ln(x) + C - my teacher is really picky

y^4 = 4*ln(x) + 4C - and i dont remember all the terms and lingos

y^4 = 4*[ln(x) + C] - integration by parts??? please help

Answer 2

Your result is fine. I'd do it slightly differently, writing

(1/4)y^4=ln|x|+ln C = (1/4)y^4=ln |Cx| (which I think is simpler and less trouble-prone).

Solving for y^4, multiply both sides of the equation by 4:

y^4=4 ln |Cx| = ln (Cx)^4

Then, solving for y, \[y=(\ln (Cx))^{\frac{ 1 }{ 4 }}\]

Answer 3

I see you wanted explanations of some of the steps involved. Let me know which ones that applies to.

Answer 4

all of them if possible, my prof doesn't mind if they are simple steps like dividing by 4, but integrating and such

Answer 5

quick Q how did the 4 go away in front of ln?

Answer 6

\[dx(x \frac{ dy }{ dx }=\frac{ 1 }{ y ^{3} })\rightarrow xdy = \frac{ dx }{ y ^{3} } commutative property\]
Rearrange so that all x components are together and all y comps are together; I belileve this is also due to the commutative property: multiply the entire equation through by (y^3)/x

Answer 7

Please type back to me what you now have...what does our equation look like?

Answer 8

y=(ln(Cx))14

Answer 9

y=(ln(Cx))^1/4

Answer 10

You're a couple of steps too far ahead. What I was asking you to do is the following:

Answer 11

x(dy/dx) = 1/y^3

x * dy = dx / y^3

y^3 * x dy = dx

y^3 dy = dx/x

y^4/4 = ln(x) + C

y^4 = 4*ln(x) + 4C

y^4 = 4*[ln(x) + C]

y = sqrt(2) * [ln(x) + C]^(1/4)

or

y = -sqrt(2) * [ln(x) + C]^(1/4)

thats what i got

Answer 12

\[(\frac{ y^3 }{ x })(x \frac{ dy }{ dx }=\frac{ 1 }{ y^3 }).\]Please do that and type back your result.

Answer 13

Sorry, I meant to ask you to multiply by \[\frac{ y^3dx }{ x }.\]

Answer 14

im lost im sorry

Answer 15

\[\frac{ y^3dx }{ x }(\frac{ dy }{ dx }=\frac{ 1 }{ y^3 })\rightarrow y^3dy=\frac{ dx }{ x }.\]

Answer 16

Make sure you understand this before we move on. Which algebraic property did we use here?

Answer 17

\[x(y ^{3}dy/x dx)=y^{3}dx/xy^{2}\]

Answer 18

commutative i believe

Answer 19

that's right. Compare your result to mine. You'll see that in mine, all x are grouped together in one term and all y together in another term. This is critically important. You are approaching the sol'n of this problem using the "separation of variables" method.

Answer 20

IF:\[y^3dy=\frac{ dx }{ x },\]
our next step is to integrate both sides so as to obtain an expression for \[\frac{ y^4 }{ 4 }.\]

Answer 21

Remember, we started out with an equation that has derivatives in it and want to end up with an equation in x and y that does NOT have derivatives in it. This is the basic thesis of Differential Equations.

Answer 22

OK with that? If not, ask questions now, please.

Answer 23

I completely get it

Answer 24

so y^4/4=ln|x|+c

Answer 25

So now we do the actual integration:\[\int\limits_{-}^{-}y^3dy=\int\limits_{-}^{-}\frac{ dx }{ x }\]

Answer 26

why dont you put dx at the end of it?

Answer 27

Use the power rule of integration to integrate the left side; on the right side we have a special case in which the appropriate result is ln |x| + C or len |x| + ln D or ln D|x}.

Answer 28

reciprocal rule for the right

Answer 29

Actually, I did: \[\int\limits_{-}^{-}\frac{ 1 }{ x }dx \] is exactly the same as what I wrote before. that's not "reciprocal rule," but rather a special case of the power rule for integration.

Answer 30

\[\int\limits_{-}^{-}\frac{ dx }{ x }= \ln |x| + c\]

Answer 31

on my paper i wrote integral sign dx/ x dx to close

Answer 32

Tell me when you're ready to move forward.

Answer 33

i am ready

Answer 34

Integrating both sides, we end up with:\[\frac{ y^4 }{ 4 }=\ln |x| + C, or \frac{ y^4 }{ 4 }=\ln |x| + \ln D= \ln D|x|\]

Answer 35

Each time I type something new, please respond with "I am ready to continue" or ask a pertinent question.

Answer 36

gotcha

Answer 37

Then:\[\frac{ y^4 }{ 4 }=\ln D|x|\]

Answer 38

WHICH can be multiplied by 4 (distributive property) to obtain:

Answer 39

"I am ready to continue"

Answer 40

\[y^4=4\ln D|x|=\ln (D|x|)^4\]

Answer 41

What do you think we need to do next?

Answer 42

fourth root everything

Answer 43

Because of the "ln" in there, that would not be appropriate.
We could leave the last result as is, or we could eliminate the "ln".

Do you know how to evaluate \[e^\ln x\]??

Answer 44

(ignore that " \ " character, please)

Answer 45

isn't that just x

Answer 46

Right. very good!

Answer 47

Therefore, if:\[y^4=\ln(D|x|)^4,\]

Answer 48

use each side of this equation as the exponent of e (raise e to each of these 2 "powers"):

Answer 49

\[\exp y^4=\exp \ln (D|x|)^4\]

Answer 50

then, on the left side, you'd have e^(y^4), and on the right side just (D|x|)^4.
End of solution.

Answer 51

Unless, of course, you want to solve for y; in that case you'd just take the 4th root of both sides of the equation I've just given you.

Satisfied with the explanations of the steps or have you further qeustions?

Answer 52

so e^y^4=D|x|^4

Answer 53

"exp" signifies the use of e as the base:

exp 5 translates to d^5.

Answer 54

so lets solve for y

Answer 55

All y ou have to do then is to take the fourth root of each side of my last equation;\[(y^4)^{1/4}=((D|x|)^4)^{1/4}\]

Answer 56

this answer is way different than the first one you wrote

Answer 57

yES, i realize that. Can YOU fix it?

Answer 58

ouch

Answer 59

no

Answer 60

Ouch is right. But great practice for you if you can identify my error and fix it. :)

Answer 61

i;m going to cry, this is my first semester of DE

Answer 62

was this all just a game to you

Answer 63

Aw, come on!
Look:

Answer 64

\[\exp y^4=\exp \ln(D|x|)^4 becomes \exp y^4=(D|x|)^4.\]

Answer 65

Personally, I'd leave the answer like that.

Answer 66

One more time: any final question or questions?

Answer 67

Then, solving for y,
y=(ln(Cx))14

Answer 68

that is what you wrote as the answer way up above

Answer 69

Certainly not just a game to me, but I'm a retired math professor and have been through these steps countless times.

Answer 70

you wrote

"Then, solving for y,
y=(ln(Cx))^1/4"

Answer 71

I'll try to answer specific questions and am sorry I ended up with an answer different from my previous one, but do hope you will go through all this on your own and make up your own mind which result is the correct one. Eventually you'll have to do all of the work on problems such as these on your own, so the more experience you get with them, the better.

Answer 72

Jamal, it does seem that you have a point. Earlier I wrote that the final answer was something like
\[y=fourth \root of \ln (D|x|)^4\], and after having done the problem again, that's the result I got.

Apologies. But at this point you 'll have to decide what YOU want to do next. Start from scratch? or move on to another problem? or?

Answer 73

im going to cry

Answer 74

Sorry. Wish I could give you a big comfort hug. But I can't, not from California, anyway.
So: Please choose one of the alternatives I've given you.

Answer 75

this is the original equation

x(dy/dx)=1/y^3

Answer 76

Yes, I'm doing the whole problem over again on paper. But before that, could you possibly look through what we've done with a critical eye and decide which part, i f any, makes sense to you? It's not just the answer that's important, but the process of arriving at it.