Question

xydy/dx = 1+y^2/1+x^2(1+x+x^2)

Answer 1

I see the equation, bur what exactly is your question. Do you want to solve for dy/dx, find the second derivative, integrate, or what?

Answer 2

differential equation

Answer 3

\[xy \frac{ dy }{ dx }=\frac{ (1+y ^{2}) }{ 1+x ^{2}(1+x +x ^{2}) }\]

Is that the right equation? It's hard to tell from the way you've written it.

Answer 4

ya i get this but i can't integrate it

Answer 5

Also, I see it's a differential equation, but that still doesn't answer my initial question. What do you want me to do with it? Show you how to solve it?

Answer 6

ya i want to know how can i solve it

Answer 7

Oh! OK! I see you answered my question. Give me a few minutes to type the solution for you.

Answer 8

what up please help me

Answer 9

\[xy \frac{ dy }{ dx }=\frac{ 1+y ^{2} }{ 1+x ^{2}(1+x +x ^{2}) }\]
Rearrange to obtain
\[\int\limits_{}^{}\frac{ y }{ 1+y ^{2} }dy =\int\limits_{}^{}\frac{ dx }{ x[1+x ^{2}(1+x +x ^{2})] }\]

OK! I'm gonna need a few minutes to think here so be patient. Thanks.

Answer 10

\[xy \frac{ dy }{ dy }= \frac{ 1+y^2 }{ 1+x^2}(1+x+x^2)\] this is question

Answer 11

When I asked you earlier, why did you say otherwise. Could have saved both of us time. OK! so now that you've finally given me the correct equation, wait because it takes time to type a good proper solution.

Answer 12

Oh my gosh! I can't believe this. I was at the end of the solution and I had a problem with my computer. Now I have to start over again. It'll take me another ten minutes to type it all out again. Just wait.

Answer 13

ok

Answer 14

take x terms together and y terms together
\[\int\limits \frac{ y dy }{1 + y ^{2} } = \int\limits \frac{ dx }{x(1 + x^{2}) }(1+x+x ^{2})\]

Answer 15

\[xy \frac{ dy }{ dx }=\frac{ 1+y ^{2} }{ 1+x ^{2} }(1+x +x ^{2})\]
\[\int\limits_{}^{}\frac{ y }{ 1+y ^{2} }dy =\frac{ 1+x +x ^{2} }{ x(1+x ^{2}) }dx\]
It would be easier here to integrate both sides separately.

First \[\int\limits_{}^{}\frac{ y }{ 1+y ^{2} }dy\]

Using the substitution method, let u = 1 + y², then (1/2)du = ydy. Thus
\[\int\limits_{}^{}\frac{ y }{ 1+y ^{2} }dy\]
\[=\frac{ 1 }{ 2 } \int\limits_{}^{}\frac{ du }{ u }\]
\[=\frac{ 1 }{ 2 }\ln u\]
\[=\frac{ 1 }{ 2 }\ln (1+y ^{2})\]

Second \[\int\limits_{}^{}\frac{ 1+x +x ^{2} }{ x(1+x ^{2}) }dx\]
\[=\int\limits_{}^{}\frac{ 1+x ^{2} }{ x(1+x ^{2}) }dx+\int\limits_{}^{}\frac{ x }{ x(1+x ^{2}) }dx\]
\[=\int\limits_{}^{}\frac{ 1 }{ x }dx +\int\limits_{}^{}\frac{ 1 }{ 1+x ^{2} }dx\]
\[=\ln x +\tan^{-1} x +C\]

Now equate the first with the second. Thus
\[\frac{ 1 }{ 2 }\ln (1+y ^{2})=\ln x +\tan^{-1} x +C\]

Do you get it?

Answer 16

Well Vikash, I'm still waiting for a response from you.

Answer 17

thanks i properly understand