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

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

Differential Equations
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are you supposed to find dy/dx?
yup..
im not confident of the answer in my book

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Other answers:

Just re-arrange. Take the -2dy to the other side, divide both sides by dx, and then divide both sides by 2.
can you show solution??
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:\[\bf \therefore \ \frac{ xy^3\ln(x) }{ 2 }=\frac{dy}{dx}\]
variable seperable method?
@niah2411 Do you see what I did?
yes
maybe my book is wrong
add 2dy to both sides then divide both sides by 2dx.
our methods are the same and also the answers, the book answer is maybe wrong
May be. It often happens that the publishers make a mistake in the solutions.
the book answer is x^2(lnx^2-1)+4y^-2=c
That looks like you're supposed to find the solution to the equation, then, not just find dy/dx
Can you stop tagging everyone. Please don't pay attention guys.
@agent0smith can you give me a solution if you have one?
Sorry, don't really remember much of this part of calculus :(
@genius12 can you give me a solution using variable separable method?
oh..i see i understand that
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.
ohh can you give me your full solution so i can understand it really..pls?
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.
oh im sorry @agent0smith for the inconvenience
@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: \[\bf -\frac{1}{2y^2}=\frac{ 2x^2\ln(x)-x^2 }{ 8 }+C\]
Can you solve for 'y' now?
@agent0smith You still here? Help me awaken @niah2411 ? lol
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.
@agent0smith I don't know what you're saying?
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.
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
Yep, that's what i worked out. The book/yours is the same.

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