Question

\[\frac{\text dy}{\text dx}=\frac{y\left(1-y^2\right)}{x\left(1-x^2\right)}\]

Answer 1

\[\frac{\text dy}{\text dx}=\frac{y\left(1-y^2\right)}{x\left(1-x^2\right)}\]\[\frac{\text dy}{y\left(1-y^2\right)}=\frac{\text dx}{x\left(1-x^2\right)}\]\[\int\frac{\text dy}{y\left(1-y^2\right)}=\int\frac{\text dx}{x\left(1-x^2\right)}\]
\[{\frac1{z(1-z^2)}=\frac1{z(1+z)(1-z)}=\frac Az+\frac B{1+z}+\frac C{1-z}}\]\[1=({1+z})({1-z})A+z({1-z})B+z({1+z})C\]\[1=(1-z^2)A+({z-z^2})B+({z+z^2})C\]\[1=A+z(B+C)+z^2(C-A-B)\]\[1=A;\qquad B+C=0;\qquad C-A-B=0\]\[\qquad\qquad B=-C;\qquad 2C-1=0\]\[B=-1/2\qquad C=1/2;\qquad \]\[\frac1{z(1-z^2)}=\frac 1z-\frac 1{2(1+z)}+\frac 1{2(1-z)}\]
\[\int\left(\frac{1}{y}-\frac{1}{2(1+y)}+\frac{1}{2(1-y)}\right)\cdot \text dy=\int\left(\frac 1x-\frac 1{2(1+x)}+\frac 1{2(1-x)}\right)\cdot{\text dx}\]\[\ln y-\frac 12\ln(1+y)+\frac12\ln(1-y)=\ln x-\frac 12\ln(1+x)+\frac12\ln k(1-x)\]\[\ln y^2-\ln(1+y)+\ln(1-y)=\ln x^2-\ln(1+x)+\ln k(1-x)\]\[\ln \frac{y^2(1-y)}{{1+y}}=\ln \frac{x^2{(1-x)}}{{k(1+x)}}\]

Answer 2

i must have made some mistake cause the correct answer should be \[ky^2(1-x^2)=x^2(1-y^2)\]

Answer 3

oh i think i see it now

Answer 4

partial fractions seem correct to me so far.

Answer 5

you multiplied everything by 2 in the 2nd last row right?

Answer 6

yeah, , and i ve found the mistake

Answer 7

ok, it's with the multiplication of the logs?

Answer 8

should finish like this .;
\[\int\left(\frac{1}{y}-\frac{1}{2(1+y)}+\frac{1}{2(1-y)}\right)\cdot \text dy=\int\left(\frac 1x-\frac 1{2(1+x)}+\frac 1{2(1-x)}\right)\cdot{\text dx}\]
\[\ln y-\frac 12\ln(1+y)\color{red}-\frac12\ln(1-y)=\ln x-\frac 12\ln(1+x)\color{red}-\frac12\ln k(1-x)\]\[\ln y^2-\ln(1+y)-\ln(1-y)=\ln x^2-\ln(1+x)-\ln k(1-x)\]\[\ln \frac{y^2}{1-y^2}=\ln \frac{x^2}{k(1-x^2)}\]\[ky^2(1-x^2)=x^2(1-y^2)\]

Answer 9

hmm but C= (1/2)? Maybe I am overseeing something now, but glad it worked out.

Answer 10

\[\int\limits_{}^{}\frac{1}{1-y}dy=-\ln(1-y)\] ?!

Answer 11

that's it

Answer 12

thanks

Answer 13

you're welcome :)