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.

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- Jun 30th 2009, 04:00 PM #1

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- Jun 30th 2009, 05:18 PM #2
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.

- Jun 30th 2009, 05:58 PM #3

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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

- Jun 30th 2009, 06:23 PM #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 $ ?

- Jun 30th 2009, 07:48 PM #5

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- Jun 30th 2009, 08:30 PM #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

- Jun 30th 2009, 08:33 PM #7

- Jul 1st 2009, 12:58 AM #8

- Jul 1st 2009, 02:02 AM #9

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