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Solve this differential equation: $$\Biggr(\frac{dy}{dx} -1\Bigg)\Bigg(y-x\frac{dy}{dx}\Bigg)=\frac{dy}{dx}$$

How to solve this?

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$\Biggr(\frac{dy}{dx} -1\Bigg)\Bigg(y-x\frac{dy}{dx}\Bigg)=\frac{dy}{dx}\implies(p-1)(y-xp)=p$, where $p\equiv \frac{dy}{dx}$

$\implies y-xp=\frac{p}{p-1}=\implies y=xp+\frac{p}{p-1}$ . . . . $(1)$

which is known as Clairaut’s equation.

So the general solution of the differential equation $(1)$ is

$y=cx+\frac{c}{c-1}$, where $c$ is the integrating constant.


If you once show that a differential equation is of Clairaut’s form i.e., of the form $$y=px+f(p)$$

Then for general solution of this kind of ODE, just replace $p\equiv \frac{dy}{dx}$ by the integrating constant $c$. i.e., its general solution is $$y=cx+f(c)$$

For more details about the solution of Clairaut’s equation, you may follow the reference: "http://www.library.gscgandhinagar.in/assets/admin/images/MAT-102(UNIT1,2).pdf"

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    $\begingroup$ I think if $y=x+1\pm2\sqrt{x}$ is a solution of the given differential equation then it must be the singular solution. A solution of a differential equation in which the number of arbitrary constants is equal to the order of the differential equation is called the general solution or complete integral or complete primitive. A solution which can not be obtained from a general solution is called singular solution (See the reference I provide here). $\endgroup$ – nmasanta May 24 at 9:29
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    $\begingroup$ OK. That's a matter of vocabulary. So, according to the provided definitions I agree with your answer. The singular solution $y=x+1\pm2\sqrt{x}$ is the envelope of the set of straight lines $y=cx+\frac{c}{c-1}$. $\endgroup$ – JJacquelin May 24 at 9:37
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We have $y(x)=px+\frac{p}{-1+p}$. Where $p=\frac{dy}{dx}$ This is the Clairaut's equation . The solution will he of the form $y=cx+f(c)$ .Differentiating we have $dy=pdx+xdp+\frac{dp}{(1-p)^2}$ we can write $dy=pdx$ the equation becomes $-xdp=\frac{dp}{(1-p)^2}$ assuming $dp\neq 0$ we have $x=\frac{1}{(1-p)^2}$ put this value in the first equation and get an expression for $y(p)$. The $x(p),y(p) $ are called singular solutions.

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  • $\begingroup$ Shouldn't be x=1/(1-p)^2 ? $\endgroup$ – Iman Virk May 24 at 6:54
  • $\begingroup$ Corrected it. Thanks for pointing g it out. $\endgroup$ – Archis Welankar May 24 at 6:58
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Since nmasanta already posted the answer I don't edit my calculus.

I only add the singular solution $y=x+1\pm 2\sqrt{x}$ and a graph of the complete solutions of the ODE :

enter image description here

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    $\begingroup$ One should also not forget the composite solutions. One could, for example, come from the left along the $c=-0.2$ line, move some time along the singular solution and then leave to the right along the $c=0.5$ line. $\endgroup$ – LutzL May 24 at 12:12
  • $\begingroup$ Of course. You are right they are an infinity of composite solutions. $\endgroup$ – JJacquelin May 24 at 13:07
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Hint: Write $(y-x)'\left(\dfrac{x}{y}\right)'=\dfrac{y'}{y^2}$ and with substitutions $u=y-x$ and $\dfrac{x}{y}=v$ then $$\dfrac{dv}{1-v}=\dfrac{du}{u^2-u}$$

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  • $\begingroup$ $(y-x)'\left(\dfrac{x}{y}\right)'=u'v'=\frac{y'}{y^2}$ is correct. But you cannot integrate $u'v'$ as you did. $$\dfrac{du}{u}=\dfrac{dv}{v(v-1)}\quad\text{is false}$$. $\endgroup$ – JJacquelin May 24 at 7:58
  • $\begingroup$ Sorry, I don't agree. I think that $$\dfrac{dv}{1-v}=\dfrac{du}{u^2-u}\quad\text{is still false}$$ You should edit the steps of your calculus from $u'v'=\frac{y'}{y^2}$ to the proposed $\dfrac{dv}{1-v}=\dfrac{du}{u^2-u}$. $\endgroup$ – JJacquelin May 24 at 8:27
  • $\begingroup$ You should get $y=\frac{u}{1-v}$ and thus $u'v'=-(\frac1y)'=\frac{uv'-u'(v-1)}{u^2}$. Now explain how the variables get separated? $\endgroup$ – LutzL May 24 at 12:09

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