Question

help :S

Answer 1

\[y(x)=1+\int\limits_{ 0}^{x }\frac{ (y(t))^2 }{ 1+t^2 }dt\]

Answer 2

I got \[y(x)=-\frac{ 1 }{ (\arctan(x)+c) }\]
but how do I get rid of c?

Answer 3

Unfortunately, I havent seen this type of an integral before with y(t) being a part of the integral, but why would you have c in the first place? Is this not a definite integral?

Answer 4

\[\frac{ dy }{ dx }=0+\frac{ d }{ dx }\int\limits_{0}^{x}\frac{ (y(t))^2 }{ 1+t^2 }dt\]
\[\frac{ dy }{ dx }=\frac{(y(x))^2 }{ 1+x^2 }\]
\[\int\limits_{ }^{ }1/(y(x))^2=\int\limits_{ }^{ }\frac{ 1 }{ 1+x^2 }\]
\[-1/y=\arctan(x)+c\]

Answer 5

Oh, you were differentiating?

Answer 6

Edit:
\[\int\limits_{ }^{ }1/(y(x))^2dy=\int\limits_{ }^{ }\frac{ 1 }{ 1+x^2 }dx\]

Answer 7

Okay, I see what you were doing. Interesting for sure. Well, given that's the way it would be done, is there a need to eliminate c? Like is it required as a part of what you are doing, or do you just want it to go away?

Answer 8

it should have 1 answer, not infinite answers

Answer 9

got it :D

Answer 10

May I ask what you did?

Answer 11

\[y(0)=1+\int\limits_{0}^{0}........dx\]

Answer 12

I see. Basically is a built in initial condition I suppose. Cool :)

Answer 13

1=-1/(arctan(0)+c)
1=-1/(0+c)
c=-1

Answer 14

:D

Answer 15

Good job then :) Always good when you can figure it out on your own.