My aproach:

Let's make the quadratic term with other $y'$ whole to the square as follows:

$$\left(\dfrac{dy}{dx}\right)^2-4x\dfrac{dy}{dx}+4y=0\\\equiv \\ \left(\dfrac{dy}{dx}-2x\right)^2=4x^2-4y\\ \equiv \\ \dfrac{dy}{dx}=\sqrt{4x^2-4y}+2x$$

Im stuck here, what are the methods of solving these kind of nonlinear ode s? and How to continue from the last line? Thank you.


2 Answers 2


Basically this would be very hard to solve if it weren't for the specific choice of coefficients. Observe that it can be rewritten as $$ \frac{dy}{dx}-2x=\sqrt{4x^2-4y}\\ \frac{d}{dx}(y-x^2)=\sqrt{4x^2-4y}\\ -\frac{dz}{dx}=2\sqrt{z} $$ For $z=x^2-y$. Can you take it from here?

  • $\begingroup$ ohhh i feel so stupid not to see this :( thanks a lot. $\endgroup$ Nov 11, 2019 at 6:46

Alternative approach:

Given equation can be written as$$\left(\frac{dy}{dx}\right)^2-4x\frac{dy}{dx}+4y=0$$ $$\implies y=x\frac{dy}{dx}-\frac{1}{4}\left(\frac{dy}{dx}\right)^2\tag1$$ which is Clairaut's equation $\left[y(x)=x\frac{dy}{dx}+f\left(\frac{dy}{dx}\right)\right]~$.

Hence the general solution is $$y=A~x~-~\frac{1}{4}A^2$$where $~A~$ is integrating constant.

Derivation: Differentiating equation $(1)$ with rwspect to $x$, $$\frac{dy}{dx}=\frac{dy}{dx}+x~\frac{d^2y}{dx^2}-\frac{1}{2}~\frac{dy}{dx}~\frac{d^2y}{dx^2}$$ $$\implies \frac{d^2y}{dx^2}\left(x-\frac{1}{2}~\frac{dy}{dx}\right)=0$$ which gives $$\frac{d^2y}{dx^2}=0\implies \frac{dy}{dx}=A=\text{constant}$$ So general solution of equation $(1)$ is $$y=A~x~-~\frac{1}{4}A^2$$where $~A~$ is integrating constant.

  • From the remaining part we get the singular solution, $$x-\frac{1}{2}~\frac{dy}{dx}=0$$ $$\implies \frac{dy}{dx}=2x$$ Integrating $$y=x^2~.$$
  • $\begingroup$ +1 nice solution. Note that $B=0$ $\endgroup$ Nov 11, 2019 at 11:34
  • 1
    $\begingroup$ Thanks a lot @Isham $\endgroup$
    – nmasanta
    Nov 11, 2019 at 11:49

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