2y + y*x^2 - 2x = 0

We will try to free y by itself on the left side:

2y + y*x^2 = 2x

Now factor y:

y ( 2+ x^2) = 2x

y= 2x/(2+x^2)

Now let us differentiate:

y= u/v such that:

u = 2x ==> u' = 2

v= 2+x^2 ==> v' = 2x

Then,

dy = (u'v- uv')/v^2

= (2*(2+x^2) - (2x)(2x)]/(2+x)^2

= ( 4+ 2x^2 - 4x^2) / (2+x)^2

= (4- 2x^2) / (2+x)^2

Now dy's zeros"

dy = 0 when (4-2x^2) = 0

==> 2x^2 = 4

==> x^2 = 4/2 = 2

==> x= +- sqrt(2)

To verify if the given roots are the roots of dy = 0, we'll have to differentiate the given function.

We'll try to isolate y to the left side. For this reason, we'll add 2x both sides:

2y+y*x^2 = 2x

Now, we'll factorize by y:

y(x^2 + 2) = 2x

We'll divide by (x^2 + 2) both sides:

y = 2x / (x^2 + 2)

Because the function is a ratio, we'll calculate it's derivative using the quotient rule:

(u/v)'= (u'*v-u*v')/v^2

dy = [2x/(x^2+2)]'=[(2x)'*(2+x^2)-2x*(2+x^2)']/(2+x^2)^2

dy = (2x^2+4-4x^2)/(2+x^2)^2

dy = (4-2x^2)/(1+x^2)^2

We have, at numerator, a difference of squares:

a^2-b^2=(a-b)(a+b)

(4-2x^2) = (2-x*sqrt2)(2+x*sqrt2)

Because the denominator of dy is always positive, for any value of x, only the numerator could be zero.

(2-x*sqrt2)(2+x*sqrt2)=0

We'll set each factor as zero.

2-x*sqrt2=0,

x*sqrt2=2

We'll divide by sqrt2:

x = 2/sqrt2

x = 2*sqrt2/2

**x1 = sqrt2**

2+x*sqrt2 = 0

x*sqrt2 = -2

We'll divide by sqrt2:

x = -2/sqrt2

**x2 = -sqrt2**

**The roots of dy = 0 are : {-sqrt2 ; sqrt2}.**

To verify if +sqrt2 and -sqrt2 are the roots of 2y+y*x^2-2x = 0

Solution:

Rewrite the given equation as:

y*x^2 - 2x+2y = 0

Put x= sqrt2 in the equation.Then the LHS is

y (sqrt2)^2 +sqrt2 +2y = y*2 -sqrt2 +2y = -sqrt2 is not equal to RHS which is zero.so sqrt2 is not a root of the equation.

Similarly put x= -sqrt2 in the given equation. Then the LHS is:

y(-sqrt2)^2-(-sqrt2) +2y

2y +sqrt2-2y

sqrt2 but RHS value is zero.

So the sqrt2 or -sqrt2 is not verifying the given equation. So they are not the roots of the given equation.