Question

Solve the DE:
\[x\left(1-x^2\right)\text dy=\left(x^2-x+1\right)y\text dx\]

Answer 1

\[x\left(1-x^2\right)\frac{\text dy}{\text dx}=\left(x^2-x+1\right)y\]\[\frac{\text dy}{\text dx}-\frac{\left(x^2-x+1\right)}{x\left(1-x^2\right)}y=0\]

\[\mu(x)=\text{exp}\left({\int \frac{x-x^2-1}{x\left(1-x^2\right)}\text dx}\right)\]\[=\text{exp}\left({\int \frac{1}{1-x^2}-\frac{x}{1-x^2}-\frac{1}{x(1-x^2)}\text dx}\right)\]\[=\text{exp}\left(\frac 12\int\frac{2}{1-x^2}-\frac{2x}{1-x^2}-\frac{2}{x(1-x^2)}\text dx\right)\]

Answer 2

\[\int\limits_{}(x ^{2}-x+1)/(x(1-x ^{2}))dx=Log(x)-Log(1+x)-Log(1-x ^{2})/2\]

Answer 3

btw how do you type fractions?

Answer 4

\frac{a}{b} or { a \over b}
\[\frac ab\qquad {a \over b}\]

Answer 5

moving on miu(t)=x/((1+x)(1-x2))

Answer 6

\[\mu(t)=\frac{x}{(1+x)\sqrt{1-x ^{2}}}\]

Answer 7

logs are
\ln_b\left|x\right|
\[\ln_b\left|x\right| \]or \log_b\left|x\right|\[ \log_b\left|x\right|\]

Answer 8

\[y(t)=\frac{C1}{\mu(t)}\]

Answer 9

how did you get your first line/

Answer 10

wait a sec you want to do\[\mu(t)=e ^{\int\limits_{}-\frac{x ^{2}-x+1}{x(1-x ^{2})}}=\frac{(1+x)\sqrt{1-x ^{2}}}{x}\]You forgot the minus

Answer 11

\[\int\limits_{}-\frac{x ^{2}-x+1}{x(1-x ^{2})}\text dx=\int \frac{x-x^2-1}{x\left(1-x^2\right)}\text dx\]

Answer 12

\[dy/dx +d \mu/dx=0\rightarrow d(\mu \times y)/dx=0\]

Answer 13

im not sure what you have done

Answer 14

well in that case i forgot the minus

Answer 15

i will try to write it down give me a sec

Answer 16

\[\frac{1}{2}\int\limits{}\frac{-2x}{1-x ^{2}}=\frac{Log(1-x ^{2}}{2}\]That's the easiest one, moving on...

Answer 17

sorry, had to go for breakfast,
\[\frac{x-x ^{2}-1}{x(1+x)(1-x)}=\frac{A}{x}+\frac{B}{1+x}+\frac{C}{1-x}\rightarrow x-x ^{2}-1=A(1-x ^{2})+Bx(1-x)+Cx(1+x)\]
\[(-A-B+C)x ^{2}+(B+C)x+A=x-x ^{2}-1\]\[A=-1\]\[B+C=1\]\[1-B+C=-1,C=1-B \rightarrow 1- B +1 - B=1->B=3/2,C=-1/2\]
\[A=-1\]\[B=\frac{3}{2}\]\[C=-\frac{1}{2}\]
We get\[-\frac{1}{x}+\frac{3}{2+2x}+\frac{1}{2x-2}\]

Answer 18

solving,\[\int\limits_{}-\frac{1}{x}dx=-Log(x)\]\[\int\limits_{}\frac{3}{2+2x}dx=\frac{3Log(2+2x)}{2}\]
\[\int\limits_{}\frac{1}{2x-2}dx=\frac{Log(-2+2x)}{2}\]

Answer 19

\[\frac{3Log(2+2x)+Log(-2+2x)}{2}=\frac{Log[(2+2x)^{3}(-2+2x)]}{2}=\]\[\frac{Log[(2+2x)^{2}(4x ^{2}-4)]}{2}=\frac{Log[\frac{(x+1)^{2}}{4}4(x ^{2}-1)]}{2}=\frac{Log[(x+1)^{2}(x ^{2}-1)]}{2}\]\[Log(x+1)+\frac{Log(x ^{2}-1)}{2}\]

Answer 20

Thus,
\[\int\limits \frac{x-x ^{2}-1}{x(1-x ^{2})}dx=-Log(x)+Log(x+1)+\frac{Log(1-x ^{2})}{2}\]

Answer 21

\[x\left(1-x^2\right)\text dy=\left(x^2-x+1\right)y\text dx\]\[x\left(1-x^2\right)\frac{\text dy}{\text dx}=\left(x^2-x+1\right)y\]
\[\frac{\text dy}{\text dx}-\frac{\left(x^2-x+1\right)}{x\left(1-x^2\right)}y=0\]\[\mu(x)=\text{exp}\left({\int \frac{x-x^2-1}{x\left(1-x^2\right)}\text dx}\right)\]\[=\text{exp}\left({\int \frac{1}{1-x^2}-\frac{x}{1-x^2}-\frac{1}{x(1-x^2)}\text dx}\right)\]\[=\text{exp}\left(\frac 12\int\frac{2}{1-x^2}-\frac{2x}{1-x^2}-\frac{2}{x(1-x^2)}\text dx\right)\]\[=\text {exp}\left(\frac 12\left(\ln|x+1|-\ln|1-x|+\ln\left|1-x^2\right|+\ln|1-x^2|-2\ln|x| \right)\right)\]\[=\text {exp}\left(\frac 12( \ln\left|\frac{(x+1)(1-x^2)(1-x^2)}{(1-x)x^2}\right|)\right)\]\[=\text {exp}\left(\frac 12\ln\left|\frac{(x+1)(1-x^2)^2}{(1-x)x^2}\right|\right)\]

\[=\text {exp}\left(\ln\left|\left(\frac{(x+1)(1-x^2)^2}{(1-x)x^2}\right)^{1/2}\right|\right)\]\[=\text {exp}\left(\ln\left|\sqrt{\frac{(x+1)(1-x^2)^2}{(1-x)x^2}}\right|\right)\]\[\mu(x)=\left|\sqrt{\frac{(x+1)(1-x^2)^2}{(1-x)x^2}}\right|\]
\[\frac{\text d}{\text dx}(\mu y)=0\]\[\mu y=\int0\text dx\]\[\mu y=c\]\[y=\frac c \mu\]\[y=\frac{c}{\left|\sqrt{\frac{(x+1)(1-x^2)^2}{(1-x)x^2}}\right|}\]\[y=\frac{cx\sqrt{(1-x)}}{\sqrt{(x+1)}(1-x^2)}\]\[y=\frac{cx}{\sqrt{x+1}\sqrt{1-x}(1+x)}\]\[y=\frac{cx}{\sqrt{1-x^2}(1+x)}\]