# Nonlinear differential equation $\left(\frac{dy}{dx}\right)^2-4x\frac{dy}{dx}+4y=0$

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.

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?

• ohhh i feel so stupid not to see this :( thanks a lot. Commented 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 $$y=x^2~.$$
• +1 nice solution. Note that $B=0$ Commented Nov 11, 2019 at 11:34
• Thanks a lot @Isham Commented Nov 11, 2019 at 11:49