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

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

How to solve this?

$$\Biggr(\frac{dy}{dx} -1\Bigg)\Bigg(y-x\frac{dy}{dx}\Bigg)=\frac{dy}{dx}\implies(p-1)(y-xp)=p$$, 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 $$y=cx+f(c)$$

• 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). – nmasanta May 24 at 9:29
• 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 $y=cx+\frac{c}{c-1}$. – JJacquelin May 24 at 9:37

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.

• Shouldn't be x=1/(1-p)^2 ? – Iman Virk May 24 at 6:54
• Corrected it. Thanks for pointing g it out. – Archis Welankar May 24 at 6:58

I only add the singular solution $$y=x+1\pm 2\sqrt{x}$$ and a graph of the complete solutions of the ODE :
• 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. – LutzL May 24 at 12:12
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 $$\dfrac{dv}{1-v}=\dfrac{du}{u^2-u}$$
• $(y-x)'\left(\dfrac{x}{y}\right)'=u'v'=\frac{y'}{y^2}$ is correct. But you cannot integrate $u'v'$ as you did. $$\dfrac{du}{u}=\dfrac{dv}{v(v-1)}\quad\text{is false}$$. – JJacquelin May 24 at 7:58
• 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 $\dfrac{dv}{1-v}=\dfrac{du}{u^2-u}$. – JJacquelin May 24 at 8:27
• 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? – LutzL May 24 at 12:09