Question

How to solve this differential equation:
2 dy/dx = 1-y*2

Answer 1

Yes.

Answer 2

\[2 \frac{\text dy}{\text dx} = 1-y^2\]
\[ \frac{2\text dy}{1-y^2} = \text dx\]

Answer 3

most likely...

Answer 4

i dont get it

Answer 5

partial fractions?

Answer 6

break up the square?

Answer 7

\[\frac{2}{1-y^2}=\frac A{1+y}+\frac B{1-y}\]yeah

Answer 8

I don't know how to do it.

Answer 9

Can someone please show me how to answer this question.

Answer 10

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

Answer 11

\[A+B=2\]\[B-A=0\]

Answer 12

A=B=1

Answer 13

I know how to do the partial fraction part but I'm stuck after that.

Answer 14

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

Answer 15

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

Answer 16

\[\ln|1+x|-\ln|1-y|=x+c\]

Answer 17

*\(\int dx\)

Answer 18

\[\ln\left|\frac{1+x}{1-y}\right|=x+c\]
\[\frac{1+y}{1-y}=e^ce^x\]
\[\frac{1+y}{1-y}=Ae^x\]

Answer 19

*\(\ln|1+y|\)

Answer 20

In your third step, why did you bring the minus sign out. why the (+) sign can't remain ?

Answer 21

\[u=1-y\implies du=-dy\]he just wrote it in such a way as to not make you do the -substitutiuon

Answer 22

u-substitution*

Answer 23

it is a strange step , and i dont really get it my self, if you try to integrate with out the negative you get arc-tans instead of logs

Answer 24

Yes, that's what I mean. If the (+) sign remains then the answer will be totally different, Anyway thanks.

Answer 25

i think it might have something to do with the domain of x