y-2-xy-y-0-find-the-solution-

Question Number 87059 by jagoll last updated on 02/Apr/20

{(y^{'})}^{2} - {xy}^{'} + y = 0

find the solution

Answered by mr W last updated on 03/Apr/20

y^{'} = \frac{1}{2} (x \pm \sqrt{x^{2} - 4 y})

l e t u = x^{2} - 4 y

c a s e 1 : u = c o n s t a n t = C

\Rightarrow x^{2} - 4 y = - 4 C

\Rightarrow y = \frac{x^{2}}{4} + C

y^{'} = \frac{x}{2}

{(\frac{x}{2})}^{2} - x (\frac{x}{2}) + \frac{x^{2}}{4} + C = 0 \Rightarrow C = 0

\Rightarrow y = \frac{x^{2}}{4}

c a s e 2 : u \neq c o n s t a n t

\frac{d u}{d x} = 2 x - 4 y^{'}

2 x - \frac{d u}{d x} = 2 x \pm 2 \sqrt{u}

\frac{d u}{d x} = \pm 2 \sqrt{u}

\int \frac{d u}{2 \sqrt{u}} = \pm \int d x

\sqrt{u} = C \pm x

u = {(C \pm x)}^{2} = x^{2} + C (C \pm 2 x)

x^{2} - 4 y = x^{2} + C (C \pm 2 x)

\Rightarrow y = - \frac{C (C \pm 2 x)}{4}

c h e c k :

y^{'} = - \frac{\pm C}{2}

{x y}^{'} = - \frac{\pm C x}{2}

{x y}^{'} - y = - \frac{\pm 2 C x}{4} + \frac{C (C \pm 2 x)}{4} = \frac{C^{2}}{4} = {(y^{'})}^{2}

Commented by jagoll last updated on 03/Apr/20

sir y = \frac{x^{2}}{4} ?

Commented by mr W last updated on 03/Apr/20

y e s . y = \frac{x^{2}}{4} i s a l s o a s o l u t i o n, s e e

a b o v e .

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