1. ## another question!

I am having problem solving this problem. Kind of stuck. Help would be veryy appreciated.
the problem is :
The differential equation y' = x^2 - y^2 -1 has y = -x as one of its solutions find the another solution.

2. Originally Posted by blazingwes
I am having problem solving this problem. Kind of stuck. Help would be veryy appreciated.
the problem is :
The differential equation y' = x^2 - y^2 -1 has y = -x as one of its solutions find the another solution.
Your DE has the general form $\displaystyle \frac{dy}{dx} = p(x) + q(x) y + r(x) y^2$ (known as a Riccati equation).

The usual technique for this type of DE is to make the substitution $\displaystyle z = \frac{1}{y - f(x)}$ where $\displaystyle y = f(x)$ is a particular solution to the DE.

3. i used one of the solution and
i let y = u - x
therefore y' = u'-1

plug it back into the equation and i got
u'-1 = x^2 - ( u-x)^2 -1
u'-1 = x^2 - (u^2-2ux+x^2)-1
u'-1 = x^2 - u^2 + 2ux - x^2 -1
u' = -u^2+2ux
u' = u(-u+2x)

I dunno if i am doing it correctly or not... i am stuck at the integration part

4. I ran into a problem.

$\displaystyle \frac {dy}{dx} = x^{2} -y^{2} -1$

Doing as mr fantastic suggested, let $\displaystyle z = \frac{1}{y+x}$

then $\displaystyle y = \frac {1}{z} - x$

$\displaystyle \frac {dy}{dx} = \frac {\text{-}1}{z^{2}} \frac {dz}{dx} - 1$

$\displaystyle \frac {\text{-}1}{z^{2}} \frac {dz}{dx} - 1 = x^{2} - \Big(\frac{1}{z} - x \Big)^{2} -1$

$\displaystyle \frac {\text{-}1}{z^{2}} \frac {dz}{dx} = \frac {\text{-}1}{z^{2}} + \frac {2x}{z}$

$\displaystyle \frac {dz}{dx} = 1 - 2xz$

$\displaystyle \frac {dz}{dx} + 2xz = 1$

now we have a first order linear differential equation

$\displaystyle e^{x^{2}} \frac {dz}{dx} + 2xz e^{x^{2}} = e^{x^{2}}$

$\displaystyle \frac {d}{dx} e^{x^{2}}z = e^{x^{2}} \ dx$

$\displaystyle e^{x^{2}}z = \int e^{x^{2}} \ dx$ ?

5. continuing from what i did.... i got the same answer too

6. The most you can do is leave the answer as $\displaystyle z = e^{\text{-}x^{2}} \Big(\int_{x_{0}}^{x} e^{t^{2}} \ dt + C \Big)$

then $\displaystyle \frac {1}{y+x} = e^{\text{-}x^{2}} \Big(\int_{x_{0}}^{x} e^{t^{2}} \ dt + C\Big)$

or $\displaystyle y = \frac {1}{e^{\text{-}x^{2}} \Big(\int_{x_{0}}^{x} e^{t^{2}} \ dt + C\Big)}- x$

yuck

7. lol yahhh yuck

8. Originally Posted by blazingwes
lol yahhh yuck
- Wolfram|Alpha

9. Originally Posted by Random Variable
I ran into a problem.

$\displaystyle \frac {dy}{dx} = x^{2} -y^{2} -1$

Doing as mr fantastic suggested, let $\displaystyle z = \frac{1}{y+x}$

then $\displaystyle y = \frac {1}{z} - x$

$\displaystyle \frac {dy}{dx} = \frac {\text{-}1}{z^{2}} \frac {dz}{dx} - 1$

$\displaystyle \frac {\text{-}1}{z^{2}} \frac {dz}{dx} - 1 = x^{2} - \Big(\frac{1}{z} - x \Big)^{2} -1$

$\displaystyle \frac {\text{-}1}{z^{2}} \frac {dz}{dx} = \frac {\text{-}1}{z^{2}} + \frac {2x}{z}$

$\displaystyle \frac {dz}{dx} = 1 - 2xz$

$\displaystyle \frac {dz}{dx} + 2xz = 1$

now we have a first order linear differential equation

$\displaystyle e^{x^{2}} \frac {dz}{dx} + 2xz e^{x^{2}} = e^{x^{2}}$

$\displaystyle \frac {d}{dx} e^{x^{2}}z = e^{x^{2}} \ dx$

$\displaystyle e^{x^{2}}z = \int e^{x^{2}} \ dx$ ?
and then use:

$\displaystyle \text{erfi}(x)=\frac{2}{\sqrt{\pi}}\int_0^x e^{t^2}\ dt$

to simplify this.

CB