Question

Obtaining general solution:
a.) ylnx(dx/dy)=[(y+1)/x]^2

b.) dy/dx=(xy+3x-y-3)/(xy-2x+4y-8)

Need help. Thanks :)

Answer 1

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Answer 2

looks like separation of variables, did u try ?

Answer 3

\[\large y \ln x \dfrac{dx}{dy}=\left(\dfrac{y+1}{x}\right)^2\]

Answer 4

like that ?

Answer 5

Ok. I'll try. Thanks! :)

Answer 6

xlnxdx=[(y+1)^2dy]/y
Is this correct??

Answer 7

\[\large x^2\ln x dx=\dfrac{(y+1)^2}{y} dy\]

Answer 8

yes integrate both sides

Answer 9

\[\large \int x^2\ln x dx=\int \dfrac{(y+1)^2}{y} dy\]

Answer 10

Thank you!!! :)

Answer 11

np :) you need to work the left hand side by parts... hope you can finish it off

Answer 12

for the second problem, factor the right hand side first :
http://www.wolframalpha.com/input/?i=factor+%28xy%2B3x-y-3%29%2F%28xy-2x%2B4y-8%29

Answer 13

good luck !

Answer 14

Did I got it right?
I obtained this equation..
1/3(x^2)(lnx-1/3)=C

Answer 15

I mean from the left side only..

Answer 16

\[\large \int x^2\ln x dx=\int \dfrac{(y+1)^2}{y} dy \]

\[\large \frac{1}{3}x^3\ln x - \frac{x^3}{9}=\int \dfrac{y^2 + 2y + 1}{y} dy \]

\[\large \frac{1}{3}x^3\ln x - \frac{x^3}{9}=\int y + 2 + \dfrac{ 1}{y} dy \]

\[\large \frac{1}{3}x^3\ln x - \frac{x^3}{9}= \dfrac{y^2}{2} + 2y + \ln y + C \]

Answer 17

for the left side you should get :
1/3(x^3)(lnx-1/3)