Question

Solve using separation methods:\[\left(y-yx^2\right)\frac{\text dy}{\text dx}=\left(y+1\right)^2\]

Answer 1

god I hate implicit diffrentiation

Answer 2

i think it is separable

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

now what can i do

Answer 7

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

Answer 8

the answer in the back of the book has log's in it

Answer 9

homogeneous!!!! woo!! lol

Answer 10

\[
\frac{y}{(y+1)^2}=\frac{1}{y+1}-\frac{1}{(y+1)^
2}
\]

Answer 11

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

Answer 12

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

Answer 13

\[\ln|y+1|+\frac 1{y+1}=\int \frac{1}{2 (x+1)}-\frac{1}{2(x-1)}
\text dx\]
\[\ln|y+1|+\frac 1{y+1}=\frac 12 \left(\ln|x+1|-\ln|x-1|\right)+c\]
\[\ln|y+1|+\frac 1{y+1}=\frac 12 \ln\left|\frac{x+1}{x-1}\right|+c\]

Answer 14

which is the answer in the the back of my book
thank you @eliassaab

Answer 15

is that factorization ment to be obvious

Answer 16

This is called decomposition in partial fractions.

Answer 17

ah yes i suppose intermediate steps would have involved A and B which turned out to be 1,1 and 1/2,-1/2

Answer 18

yes