Solution by differential equation by Clairaut Form Solve the differential equation:
$$y+x \frac{dy}{dx}=x^4 \bigg(\frac{dy}{dx}\bigg)^2$$
This is given under 'Clairaut form' but I am not able to convert it to Clairaut form of type $y=px+f(p)$ where $p=dy/dx$. The general solution is given as $xy+c=c^2x$ and singular solution is $4x^2y+1=0$. Could someone give me some hint with this?
 A: Take the derivative of the equation to obtain
$$2\frac{dy}{dx}+x \frac{d^2y}{dx^2}=4x^3 \bigg(\frac{dy}{dx}\bigg)^2+ 2x^4\frac{dy}{dx}\frac{d^2y}{dx^2}$$
Factorize the right hand of the equation
$$2\frac{dy}{dx}+x \frac{d^2y}{dx^2}=2x^3\frac{dy}{dx} \bigg(2\frac{dy}{dx}+ x\frac{d^2y}{dx^2}\bigg)$$
Now you see that you can rewrite this as
$$\bigg(1 - 2x^3\frac{dy}{dx}\bigg) \bigg(2\frac{dy}{dx}+ x\frac{d^2y}{dx^2}\bigg) = 0$$
So you have two linear ODEs that you can solve, the first giving
$$y=-\frac{1}{4x^2}+C$$
as a solution, the second giving
$$y=-\frac{A}{x}+K$$
However, when you fill them back into your original equation, you'll notice that these only satisfy it provided that
$$C=0 \; \text{ and } \; K=A^2$$
This settles the problem.
A: $$y=-x \frac{dy}{dx}+x^4 \bigg(\frac{dy}{dx}\bigg)^2$$
Let $\quad x=\frac1t\quad\to\quad dx=-\frac{dt}{t^2}\quad\to\quad \frac{dy}{dx}=\frac{dy}{dt}\frac{dt}{dx}= -t^2\frac{dy}{dt}$
$y=-\frac1t \bigg(-t^2\frac{dy}{dt}\bigg)+\bigg(\frac1t\bigg)^4 \bigg(-t^2\frac{dy}{dt}\bigg)^2$
$$y=t\frac{dy}{dt}+\bigg(\frac{dy}{dt}\bigg)^2 $$
$\frac{dy}{dt}=\frac{dy}{dt}+t\frac{d^2y}{dt^2}+2\frac{dy}{dt}\frac{d^2y}{dt^2}$
$$\bigg(t+2\frac{dy}{dt}\bigg)\frac{d^2y}{dt^2}=0$$
General solution :
$\frac{d^2y}{dt^2}=0\quad\to\quad y=at+b=\frac{a}{x}+b=-x(\frac{-a}{x^2})+x^4(\frac{-a}{x^2})^2=\frac{a}{x}+a^2\quad\to\quad b=a^2$
$$y=\frac{a}{x}+a^2$$
and particular solution :
$t+2\frac{dy}{dt}=0 \quad\to\quad y=-\frac{t^2}{4}+c=-\frac{1}{4x^2}+c=-x(\frac{1}{2x^3})+x^4(\frac{1}{2x^3})^2=-\frac{1}{4x^2}\quad\to\quad c=0$
$$y=-\frac{1}{4x^2}$$
