Question

How to solve this differential equation:
(1+x)*2 dy/dx+y*2=1

Answer 1

looks separable to me ....

Answer 2

i take it you used * instead of ^, but meant ^

Answer 3

Yes.

Answer 4

do you know how to do separable stuff?

Answer 5

Yes, but I just want to check my steps whether its correct.

Answer 6

show me what you got, ill see if its proper

Answer 7

\[f(x)\frac{dy}{dx}+g(y)=c\]
\[f(x)\frac{dy}{dx}=c-g(y)\]
\[f(x)dy=[c-g(y)]dx\]
\[\frac{dy}{c-g(y)}=\frac{dx}{f(x)}\]

is what i see it as

Answer 8

I got until here and then I'm stuck.
\[\int\limits_{}^{} 1/1+y*2 dy = \int\limits_{}^{} 1/(1+x)*2 dx\]

Answer 9

the rest is just integrating tho ...

Answer 10

your swapping things about looks good ... except that should be a -y^2 over there

Answer 11

I forgot how to integrate this type of function and I can't find any examples in my books.

Answer 12

the right side, the xs is just a power rule

the left side looks triggy to me

Answer 13

or since its a difference of squares, it might have to go .... whats the term for splitting the denom?

Answer 14

partial fraction decomp

Answer 15

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

Answer 16

\[\frac{1}{1-y^2}=\frac{a}{1-y}+\frac{b}{1+y}\]
\[1=a(1+y)+b(1-y)\]
\[1=a+ay+b-by\]
\[1=(a-b)y + (a+b)\]

a-b=0
a+b=1
------
2a = 1 ; a=1/2

b has to be -1/2

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

Answer 17

hmm, i think b = 1/2 not -1/2 :)

Answer 18

Yes, b= 1/2.

Answer 19

then after that its back to simple integration

Answer 20

\[\int(\frac{1}{2(1-y)}+\frac{1}{2(1+y)})dy\]

Answer 21

partial fraction decomp is what everyone moaned and groaned about in class last semester :)

Answer 22

I still groan and moan about it till now even though it was last semester's work cause it is related to this semester's topic.

Answer 23

Hey if y=0 and x=0 then how do I solve it ?

Answer 24

\[-\frac{1}{2}ln(1-y)+\frac{1}{2}ln(1+y)\]
\[\frac{1}{2}[ln(1+y)-ln(1-y)]\]
\[\frac{1}{2}ln(\frac{1+y}{1-y})=-\frac{1}{1+x}+C\]
\[x=y=0\]
\[\frac{1}{2}ln(1)=-1+C\]
\[0=-1+C\]
\[1=C\]

Answer 25

once youve integrated; just plug in y and x to determine "C"

Answer 26

In your fist step, how did you get\[-1/2\ln (1-y)\]. How did you get the (-) sign ? Shouldn't it be a positive value ?

Answer 27

Can someone please answer me.

Answer 28

\[u=1-y\implies du=-dy\implies dy=-du\]

Answer 29

\[\int\frac{dy}{1-y}=-\int\frac{du}u=-\ln u=-\ln|1-y|+C\]

Answer 30

Ok, thanks.

Answer 31

i got trapped by the downtime :)