Question

how to solve: x^2(y+1)dx + y^2(x-1)dy =0 ?? i know its by separable variables, but how?

Answer 1

what have you tried ?

Answer 2

ive got: \[\frac{ x ^{2}}{ (x-1) } + \frac{ y ^{2}}{ (y+1) }\]

Answer 3

but i cannot integrate since there's no formula to do it.. :/

Answer 4

i forgot to write dx and dy respectively

Answer 5

do you get this ?\[\int \frac{ x ^{2}}{ x-1 } dx= -\int \frac{ y ^{2}}{ y+1 }dy\]

Answer 6

yes!! :), so is that an integral by parts?

Answer 7

its a rational function so we should be able to integrate it easily by first dividing

Answer 8

so its: \[\int\limits_{}^{}x -x ^{2} dx =- \int\limits_{}^{}y +y ^{2} dx \] ??

Answer 9

doesnt look right

Answer 10

\[\int \frac{ x ^{2}}{ x-1 } dx= -\int \frac{ y ^{2}}{ y+1 }dy\]

\[\int \frac{ x^{2}-1+1}{ x-1 } dx= -\int \frac{ y^{2}-1+1}{ y+1 }dy\]

\[\int \frac{ x^{2}-1}{ x-1 } + \frac{ 1}{ x-1 }dx= -\int \frac{ y^{2}-1}{ y+1 } + \frac{1}{ y+1 }dy\]

\[\int \frac{(x+1)(x-1)}{ x-1 } + \frac{ 1}{ x-1 }dx= -\int \frac{ (y+1)(y-1)}{ y+1 } + \frac{1}{ y+1 }dy\]

\[\int x+1+\frac{1}{ x-1 } dx= -\int y-1+\frac{ 1}{ y+1 }dy\]

Answer 11

see if you can finish it off

Answer 12

sure! thank u so much! so now i just integrate each thing:
x^(2)/2 + x + ln(x-1) = -y^(2)/2 + y +ln (y+1) + c