Solution by differential equation by Clairaut Form

Question

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?

Answer 1

$$y=-x \frac{dy}{dx}+x^4 \bigg(\frac{dy}{dx}\bigg)^2$$ 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 :

$t+2\frac{dy}{dt}=0 \quad\to\quad y=-\frac{t^2}{4}+c=-\frac{1}{4x^2}+c=-x(\frac{1}{2x^3})+x^4(\frac{1}{2x^3})^2=-\frac{1}{4x^2}\quad\to\quad c=0$ $$y=-\frac{1}{4x^2}$$

Answer 2

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.