# $y'=$ ${-x+\sqrt{(x^2+4y)}}\over 2$ , $y(2)=-1$

Prove that the solutions $y_1$ and $y_2$ for the initial value problem

$y'=$${-x+\sqrt{(x^2+4y)}}\over 2$ , $y(2)=-1$

are:

$y_1=1-x$

$y_2={{-x^2}\over 4}$

And explain why the two solutions not contradiction with the theory of existence and uniqueness.

( sorry I don't speak English well )

For the initial condition :

$y_1=1-x$ Satisfy the initial condition

$y(2)=-1$

$y_1=1-2=-1$

And Satisfy the differential equation

${y_1}^{\prime}=-1$

${{-x+\sqrt{x^2+4y}}\over 2}={{-x+\sqrt{x^2+4(1-x)}}\over 2}$

$={{-x+\sqrt{(x-2)^2}}\over 2}$

$={{-x+x-2}\over 2}=-1$

L.h.s.=R.h.s.

$y^{\prime}={{-x+\sqrt{x^2+4y}}\over 2}$

Also $y_2$ Satisfy the initial condition

$y_2={{-x^2}\over4}={-4\over4}=-1$

• And satisfy the differential equation.

$y_2^{\prime}={-x\over 2}$

${{-x+\sqrt{x^2+4y}}\over 2}={{-x+\sqrt{x^2+4({-x^2\over 4})}}\over 2}={-x\over 2}$

And for second part of exercise:

${\partial f\over \partial y}={1\over \sqrt{x^2+4y}}$

and this not defined when $x=2, \ y=-1$

So $f(x,y)$ no satisfy lipshtize condition in any rectangular has (2,-1), Lipshtize condition including uniqueness of solution, so no contradict with the existence and uniqueness.

True ?

If this also wrong I well be delet it , I answered by the same way of my teacher :(

• Which part of the problem are you unsure about? If it's the second part, tell us why you are having trouble with it. – rogerl Oct 3 '16 at 17:23
• OP: Explain solution $y_2$: Not done in the accepted solution. "Explain why the two solutions not contradiction with the theory of existence and uniqueness": Not done in the accepted solution. Can you explain? – Did Oct 5 '16 at 8:14
• @Did I edit my answer, please is it true? Thanks. – Dima Oct 15 '16 at 21:15
• @rogerl I edit my answer, please is it true? Thanks. – Dima Oct 15 '16 at 21:15
• Sorry but what are you playing at? The new accepted solution misses $y_2$ and squarely does not "explain why the (existence of) two solutions (does) not contradict (...) existence and uniqueness". You could have stayed with the first accepted solution, it was as wrong as the present one but not more. – Did Oct 16 '16 at 8:32

Let $u=\sqrt{x^2+4y}$ ,

Then $y=\dfrac{u^2-x^2}{4}$

$\dfrac{dy}{dx}=\dfrac{u}{2}\dfrac{du}{dx}-\dfrac{x}{2}$

$\therefore\dfrac{u}{2}\dfrac{du}{dx}-\dfrac{x}{2}=\dfrac{u-x}{2}$ with $u(2)=0$

$\dfrac{u}{2}\dfrac{du}{dx}-\dfrac{x}{2}=\dfrac{u-x}{2}$ with $u(2)=0$

$\dfrac{u}{2}\dfrac{du}{dx}=\dfrac{u}{2}$ with $u(2)=0$

$\dfrac{du}{dx}=1$ or $u=0$ with $u(2)=0$

$u=x-2$ or $u=0$

$\sqrt{x^2+4y}=x-2$ or $\sqrt{x^2+4y}=0$

$x^2+4y=x^2-4x+4$ or $x^2+4y=0$

$4y=-4x+4$ or $4y=-x^2$

$y=1-x$ or $y=-\dfrac{x^2}{4}$

• This identifies one solution and misses the other, dividing carelessly by $u$. Rather amazingly this does not seem to trouble the author of the question although this same fact was already signalled to them à propos another answer, now deleted, and although the aim of the exercise is to understand this non uniqueness... – Did Oct 16 '16 at 8:29