Radical Expression Everywhere!

Square both sides :- $(y+1)+(y-1)-2\sqrt{y^2-1}=4y-1$ $\implies -2\sqrt{y^2-1}=2y-1$

Squaring again gives:-

$4(y^2-1)=(2y-1)^2$ $\implies 4y-5=0\implies y=\dfrac 54$

Substituting $\dfrac 54$ in the eqn, we get:-

$RHS:\left(\dfrac 94\right)^{1/2}-\left(\dfrac 14\right)^{1/2}=1$ While $LHS$ comes out to be $(4)^{1/2}=2$ . Hence $x=\dfrac 54$ is a extraneous solution obtained due to squaring. Hence $\color{#D61F06}{\text{No solution}}$ .

We have to solve the equation $\sqrt{y+1}-\sqrt{y-1}=\sqrt{4y-1}$ . Squaring Both sides

$\implies y+1+y-1-2\sqrt{y^2-1}=4y-1 \\ 4(y^2-1)=(2y-1)^2 \\ 4y^2-4=4y^2+1-4y \\ y=\color{#3D99F6}{\boxed{\dfrac54}}$

In this case , The option " There is no solution " is correct

On squaring, we have,
y+1 + y-1 - 2√(y^2 - 1) =(4y -1),
Or, - 2√(y^2-1) = (2y - 1),
Again squaring, we have,
4(y^2 - 1) = 4y^2 - 4y + 1,
4y = 5, or y = 5/4.
This answer 5/4 is actually an answer to only the last equation, 4y = 5, and not an answer to the original equation for the reason that same answer would be found for the following equations also:
1. - √(y+1) + √(y - 1) = √(4y - 1),
2. +√(y+1) + √(y-1) = √(4y-1),
And the same answer would also be found if the sign is changed from (+) to (- )on the RHS for each of these equations.
As such, on squaring, different equations may become identical and may generate answers which may not satisfy the original equations. Unless the answer obtained satisfies a particular equation, it can not be treated as pertaining to that equation. As y=5/4 does not satisfy the original equation in the instant case, the original equation does not have any solution.