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For the sulotion of the OED: $\frac{dy}{dx}=1+x+y^2+xy$. This is a Riccati Equation! Anyone can give me hint?

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up vote 2 down vote accepted

if we know $y_p=-\frac{x^2-2x-1}{x^2-4x+3}$ is a particular solution. So we can find general solution of Riccati Equation. (Note:I did not try it .I trust GEdgar. Thanks to him. I just offer a general solution If we know a particular solution)

Use transform of $y=y_p+\frac{1}{H} $


$y_p'+(\frac{-H'}{H^{2}})=1+x+(y_p^2+\frac{2y_p}{H}+\frac{1}{H^2})+xy_p+\frac{x}{H} $

$y_p'=1+x+y_p^2+xy_p+\frac{H'}{H^{2}}+\frac{2y_p}{H}+\frac{1}{H^2}+\frac{x}{H} $

if $y_p$ is a solution of the equation then it must satisfy $y_p'=1+x+y_p^2+xy_p$ Thus $0=\frac{H'}{H^{2}}+\frac{2y_p}{H}+\frac{1}{H^2}+\frac{x}{H} $

It is a homework , So Now you should see a linear first order differantial equation here and then can get the general solution:

$$y=y_p-\frac{e^{-\int{(2y_p+x)}dx}}{\int{e^{-(\int{2y_p+x)}dx}}dx} $$

For more information:It is not always so easy to find a particular solution of Riccati equation. I dont know a closed form method to find a particular solution. How to find general solution of Riccatti equation have been asked and offered some methods in my question

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Changing $y=u-\frac{z}{2}$ the equation is cast to the form $$\frac{du}{dx}=u^2+\left(\frac{x}{2}-1\right)^2+\frac{5}{2}$$ which does not seem to possess solutions in terms of elementary functions

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Actually, a particular solution is $$y=-\frac{x^{2} - 2 x - 1}{3 + x^{2} - 4 x}$$ and isn't there a known method to find the general solution given a particular solution? – GEdgar Jun 4 '12 at 16:20
certainly, let $u=\frac{1}{y-y_1}$ where $y_1$ is the particular integral. it's amazing how many solutions come to mind once the answer is known :) – Valentin Jun 4 '12 at 20:10 gives a simpler result. This implies that changing to an 2nd linear ODE approach should be better than using this ugly particular solution to solve. – doraemonpaul Jun 5 '12 at 1:57
You particular solution is correct. But I want to know how you get this solution. – Riemann Jun 6 '12 at 3:31

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