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

- anonymous

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

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- anonymous

are you supposed to find dy/dx?

- anonymous

yup..

- anonymous

im not confident of the answer in my book

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## More answers

- anonymous

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

- anonymous

can you show solution??

- anonymous

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:\[\bf \therefore \ \frac{ xy^3\ln(x) }{ 2 }=\frac{dy}{dx}\]

- anonymous

variable seperable method?

- anonymous

@niah2411 Do you see what I did?

- anonymous

yes

- anonymous

maybe my book is wrong

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- agent0smith

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

- anonymous

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

- anonymous

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

- agent0smith

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

- anonymous

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

- anonymous

oh..i see i understand that

- agent0smith

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.

- anonymous

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

- agent0smith

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.

- anonymous

oh im sorry @agent0smith for the inconvenience

- anonymous

@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: \[\bf -\frac{1}{2y^2}=\frac{ 2x^2\ln(x)-x^2 }{ 8 }+C\]

- anonymous

Can you solve for 'y' now?

- anonymous

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

- agent0smith

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.

- anonymous

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

- agent0smith

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.

- anonymous

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

- agent0smith

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

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