Solve $(y')^2 - xy' + y = 0$

Question

I am so confused, how to get a solution of following differential equation:

$$(y')^2 -xy' + y = 0$$

My Thought:

Try solving $y'$ to get $y' = \dfrac{ x \pm \sqrt{x^2 - 4y}}{2}$ and now its unsolvable. Kindly help, appreciated.

Answer 1

If you take the derivative of your equation, you get:

$$2y'y''-xy''-y'+y' = 0$$

or

$$y''(2y'-x)=0.$$

Let $v=y'$ and we have $v'(2v-x) = 0,$ so either $v'=0$ and $v=c$ or $v=x/2$.

Then $y'=c$ and so $y=cx+d$ or $y'=x/2$ and $y=x^2/4.$

Plugging the first into the original equation gives $d=-c^2$. So there are two solutions $y=cx-c^2$ for some constant $c$ and $y=x^2/4$. I don't know if this is all the solutions.

Answer 2

we can rewrite the O.D.E as $$(y'-\frac{x}{2})^2=-(y-\frac{x^2}{4})$$ now let

$$u=(y-\frac{x^2}{4})$$ $$u'=(y'-\frac{x}{2})$$ so $$(u')^2=-u$$ $$2u'u''=-u'$$ $$2u''=-1$$ $$u=-\frac{1}{4}x^2+C_1x+C_2$$ so the solution will be $$-\frac{1}{4}x^2+C_1x+C_2=(y-\frac{x^2}{4})$$ $$y=C_1x+C_2\tag 1$$ Plugging the equ (1) into the original equation gives $$C_2=-C_1^2$$ so the solution will be $$y=C_1x-C_1^2$$