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niah2411
 2 years ago
differential equation of xy^3lnxdx2dy=0
niah2411
 2 years ago
differential equation of xy^3lnxdx2dy=0

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genius12
 2 years ago
Best ResponseYou've already chosen the best response.3are you supposed to find dy/dx?

niah2411
 2 years ago
Best ResponseYou've already chosen the best response.0im not confident of the answer in my book

genius12
 2 years ago
Best ResponseYou've already chosen the best response.3Just rearrange. Take the 2dy to the other side, divide both sides by dx, and then divide both sides by 2.

niah2411
 2 years ago
Best ResponseYou've already chosen the best response.0can you show solution??

genius12
 2 years ago
Best ResponseYou've already chosen the best response.3First add 2dy to both sides:\[\bf xy^3\ln(x)dx2dy=0 \implies xy^3\ln(x)dx=2dy\]Now divide both sides by dx:\[\bf xy^3\ln(x)=2\frac{dy}{dx}\]Now divide both sides by 2:\[\bf \therefore \ \frac{ xy^3\ln(x) }{ 2 }=\frac{dy}{dx}\]

niah2411
 2 years ago
Best ResponseYou've already chosen the best response.0variable seperable method?

genius12
 2 years ago
Best ResponseYou've already chosen the best response.3@niah2411 Do you see what I did?

niah2411
 2 years ago
Best ResponseYou've already chosen the best response.0maybe my book is wrong

genius12
 2 years ago
Best ResponseYou've already chosen the best response.3add 2dy to both sides then divide both sides by 2dx.

niah2411
 2 years ago
Best ResponseYou've already chosen the best response.0our methods are the same and also the answers, the book answer is maybe wrong

genius12
 2 years ago
Best ResponseYou've already chosen the best response.3May be. It often happens that the publishers make a mistake in the solutions.

niah2411
 2 years ago
Best ResponseYou've already chosen the best response.0the book answer is x^2(lnx^21)+4y^2=c

agent0smith
 2 years ago
Best ResponseYou've already chosen the best response.3That looks like you're supposed to find the solution to the equation, then, not just find dy/dx

genius12
 2 years ago
Best ResponseYou've already chosen the best response.3Can you stop tagging everyone. Please don't pay attention guys.

niah2411
 2 years ago
Best ResponseYou've already chosen the best response.0@agent0smith can you give me a solution if you have one?

agent0smith
 2 years ago
Best ResponseYou've already chosen the best response.3Sorry, don't really remember much of this part of calculus :(

niah2411
 2 years ago
Best ResponseYou've already chosen the best response.0@genius12 can you give me a solution using variable separable method?

niah2411
 2 years ago
Best ResponseYou've already chosen the best response.0oh..i see i understand that

agent0smith
 2 years ago
Best ResponseYou've already chosen the best response.3Oh, 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.

niah2411
 2 years ago
Best ResponseYou've already chosen the best response.0ohh can you give me your full solution so i can understand it really..pls?

agent0smith
 2 years ago
Best ResponseYou've already chosen the best response.3Make 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.

niah2411
 2 years ago
Best ResponseYou've already chosen the best response.0oh im sorry @agent0smith for the inconvenience

genius12
 2 years ago
Best ResponseYou've already chosen the best response.3@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 rightside would just be:\[\bf \int\limits_{}^{}y^{3} \ dy=\frac{y^{2}}{2}\]Now we equate the results: \[\bf \frac{1}{2y^2}=\frac{ 2x^2\ln(x)x^2 }{ 8 }+C\]

genius12
 2 years ago
Best ResponseYou've already chosen the best response.3Can you solve for 'y' now?

genius12
 2 years ago
Best ResponseYou've already chosen the best response.3@agent0smith You still here? Help me awaken @niah2411 ? lol

agent0smith
 2 years ago
Best ResponseYou've already chosen the best response.3Hehe 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^21)+4y^2=c which you can get by moving terms around and multiplying by 8.

genius12
 2 years ago
Best ResponseYou've already chosen the best response.3@agent0smith I don't know what you're saying?

agent0smith
 2 years ago
Best ResponseYou've already chosen the best response.3This was the answer that he gave, from the book: x^2(lnx^21)+4y^2=c Which you can get from your result.

genius12
 2 years ago
Best ResponseYou've already chosen the best response.3According 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
 2 years ago
Best ResponseYou've already chosen the best response.3Yep, that's what i worked out. The book/yours is the same.
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