# Explicit differential equations

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 wolframalpha.com/input/?i=y%27^2-%28x%2By%29y%27%2Bxy%3D0 –  dato datuashvili Dec 8 '11 at 21:27
copy whole part ,not only blue –  dato datuashvili Dec 8 '11 at 21:28
so, it's good what I did? how did it get y=C1e^x? –  gov Dec 8 '11 at 21:30
see solution i have posted –  dato datuashvili Dec 8 '11 at 21:41

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 :( –  gov Dec 8 '11 at 21:45
Does the explicit example I've just added help? –  Henning Makholm Dec 8 '11 at 21:47
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. –  CogitoErgoCogitoSum Feb 22 '13 at 4:00
@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. –  Henning Makholm Feb 22 '13 at 13:28

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? –  gov Dec 8 '11 at 21:41

$y'-y=0$, 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. –  M Turgeon Dec 8 '11 at 21:51
Thank you for the tip :) I didn't know how to write correctly :) –  gov Dec 8 '11 at 22:08