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Can you help me to find all solutions of differential equation $y'^2-(x+y)y'+xy=0$?

I wrote this equation as product of explicit equations:

$$(y'-x)(y'-y)=0$$

Then I found zeroes: $y'-x=0 \Longrightarrow y'=x \Longrightarrow y=\frac{x^2}2+C_1$

$y'=0$ I don't know what to do here. Maybe to solve as equation 'without $x$'? Am I doing this right?

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    see this http://www.wolframalpha.com/input/?i=y%27^2-%28x%2By%29y%27%2Bxy%3D02011-12-08
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    copy whole part ,not only blue2011-12-08
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    so, it's good what I did? how did it get y=C1e^x?2011-12-08
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    see solution i have posted2011-12-08

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Your second equation is not $y'=0$ as you write in the question, but $y'-y=0$, in other words $y'=y$. This has the well-known solution $y=C_2e^x$.

So now you have the solutions $y=\frac{x^2}2+C_1$ and $y=C_2 e^x$. Now, for the most difficult part of the trick, you need to find all ways to glue intervals of these solutions together so the derivative matches across the glue point ... which means (consult the differential equations again!) that the gluing point(s) has to lie on the line $x=y$.

For example one solution among many would be $y=\cases{e^{x-1}&\text{for }x\le 1\\ \frac{x^2+1}2&\text{for }x\ge 1}$.

On the other hand, this gluing-together doesn't involve anything specific to differential equations, so perhaps you can take it from here?

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    Yes, I missed y in y'=y, thanks. No, I don't know what to do after I found solutions I wrote :(2011-12-08
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    Does the explicit example I've just added help?2011-12-08
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    Frankly I dont understand this talk about "glue". Why cant both solutions be valid across the entire domains? They both work everywhere, after all, if you plugged them in.2013-02-22
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    @CogitoErgoCogitoSum: Each of the "base" solutions is valid across the entire domain, but there are _additional_ solutions that can be made by using different base solutions in different intervals of the $x$ axis. So if one is interested in finding the set of _all_ solutions, as the OP wanted, then these combined solutions need to be enumerated too.2013-02-22
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The second part is $y'-y = 0 \Rightarrow y' = y \Rightarrow y = C_2e^x$

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    I've just realized that y'=y is equation that separates variables :) Thank you :) Can you look at other problem I posted?2011-12-08
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$y'-y=0$, so it means that $y'=y$ this is only if $y$ is exponential function $y=ce^x$ if $y'-x=0$ it means that $y'=x$ or $\frac{dy}{dx}=x$ integrate both we get $y=(x^2)/2+c$

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    Here's a tip: enclose your math formulas between dollar signs; it will be easier to read.2011-12-08
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    Thank you for the tip :) I didn't know how to write correctly :)2011-12-08