Question

differential equation of xy^3lnxdx-2dy=0

Answer 1

are you supposed to find dy/dx?

Answer 2

yup..

Answer 3

im not confident of the answer in my book

Answer 4

Just re-arrange. Take the -2dy to the other side, divide both sides by dx, and then divide both sides by 2.

Answer 5

can you show solution??

Answer 6

First add 2dy to both sides:\[\bf xy^3\ln(x)dx-2dy=0 \implies xy^3\ln(x)dx=2dy\]Now divide both sides by dx:\[\bf xy^3\ln(x)=2\frac{dy}{dx}\]Now divide both sides by 2:\[\bf \therefore \ \frac{ xy^3\ln(x) }{ 2 }=\frac{dy}{dx}\]

Answer 7

variable seperable method?

Answer 8

@niah2411 Do you see what I did?

Answer 9

yes

Answer 10

maybe my book is wrong

Answer 11

add 2dy to both sides then divide both sides by 2dx.

Answer 12

our methods are the same and also the answers, the book answer is maybe wrong

Answer 13

May be. It often happens that the publishers make a mistake in the solutions.

Answer 14

the book answer is x^2(lnx^2-1)+4y^-2=c

Answer 15

That looks like you're supposed to find the solution to the equation, then, not just find dy/dx

Answer 16

Can you stop tagging everyone. Please don't pay attention guys.

Answer 17

@agent0smith can you give me a solution if you have one?

Answer 18

Sorry, don't really remember much of this part of calculus :(

Answer 19

@genius12 can you give me a solution using variable separable method?

Answer 20

oh..i see i understand that

Answer 21

Oh, yes you can use separation of variables:

\[\Large \frac{ xy^3\ln(x) }{ 2 }=\frac{dy}{dx}\]
multiply both sides by dx, divide both sides by y^3\[\Large \int\limits \frac{ x\ln(x) }{ 2 } dx= \int\limits \frac{dy}{y^3}\]Now you can integrate.

Answer 22

ohh can you give me your full solution so i can understand it really..pls?

Answer 23

Make it a little simpler to integrate:\[\Large \frac{ 1 }{ 2 } \int\limits\limits x \ln(x) dx= \int\limits\limits y^{-3} dy\]
On the left it looks like you'll need to use integration by parts. Sorry, I'm too tired to go through it all now, someone else can take over.

Answer 24

oh im sorry @agent0smith for the inconvenience

Answer 25

@niah2411 I'll take over from where @agent0smith left off. To continue, we will use integration by parts for the "xln(x)" integral:
\[\bf \frac{ 1 }{ 2 }\int\limits_{}^{}xln(x) \ dx \implies \frac{ 1 }{ 2 } \left( \frac{x^2\ln(x)}{2}-\int\limits_{}^{}\frac{ x }{ 2} \ dx \right)=\frac{ 1 }{ 2 } \left( \frac{x^2\ln(x)}{2}-\frac{ x^2 }{ 4} \right)\]\[\bf =\frac{ 2x^2\ln(x)-x^2 }{ 8 }+C\]Now the right-side would just be:\[\bf \int\limits_{}^{}y^{-3} \ dy=-\frac{y^{-2}}{2}\]Now we equate the results: \[\bf -\frac{1}{2y^2}=\frac{ 2x^2\ln(x)-x^2 }{ 8 }+C\]

Answer 26

Can you solve for 'y' now?

Answer 27

@agent0smith You still here? Help me awaken @niah2411 ? lol

Answer 28

Hehe yeah I'm still here. After saying i was too tired, i did the int by parts on paper (i thought it would take longer than it did, or i would've just typed it up).

Looks like he didn't need to find y anyway, this was the answer: x^2(lnx^2-1)+4y^-2=c which you can get by moving terms around and multiplying by 8.

Answer 29

@agent0smith I don't know what you're saying?

Answer 30

This was the answer that he gave, from the book: x^2(lnx^2-1)+4y^-2=c

Which you can get from your result.

Answer 31

According to my result, which is correct, the answer should be:\[\bf x^2(2\ln(x)-1)+4y^{-2}=C\]So yes, the book answer is correct and I believe he misinterpreted the meaning of "solution" or whatever. They were just looking for separation of variables and then simply bring over the x's and y's' on one side and equating it to the constant.

@agent0smith

Answer 32

Yep, that's what i worked out. The book/yours is the same.