Solving for a matrix Y Let $X$ be a positive definite matrix and suppose that we have the expression $X = Y - Y^{-1}$, for some positive definite matrix $Y$. Can I solve this expression for $Y$ in terms of $X$?
 A: $X=Y-Y^{-1}$ implies $XY=YX$, hence $X$ and $Y$ are simultaneously diagonalizable. Thus - after changing to a suitable basis - you only have to solve the equations $x_i = y_i-y_i^{-1}$, where $x_i,y_i$ are the eigenvalues of $X$ and $Y$ respectively.
This gives you $y_i=\frac{x_i}{2} + \sqrt{\frac{x_i^2}{4}+1}$ (The other solution does not give you a positive definite matrix), which is of course the same what @Math-fun told you.
A: We will use the acronym s.p.d. for "symmetric positive definite".
Let us transform the equation into 
$$Y^2-XY-I=0 \ \ (*)$$
If we dare consider (*) as second degree matrix equation in variable $Y$, candidates for its roots (we don't say we are rigorous now; rigor is for later...) are:
$$Y_1=\frac{1}{2}(X-(X^2+4I)^{1/2}) \ \ \text{and} \ \ Y=\frac{1}{2}(X+(X^2+4I)^{1/2})$$
These writings are "legal" because


*

*$M=X^2+4I$ is s.p.d ; for proving it, we need 2 facts about s.p.d.s : a) the product of 2 s.p.d.s is an s.p.d., thus the square of an s.p.d. is an s.p.d. b) a positive shift by $4I$ preserves the fact that the matrix is symmetric; moreover, the spectrum, being "right shifted" is still $>0$. 

*Thus it makes sense to take $M^{1/2}$: the square root of a s.p.d. is a well defined concept ($M=Q \Delta Q^{-1} \ \Rightarrow \ M^{1/2}:=Q\Delta^{1/2}Q^{-1}$).
It suffices now to check that our "work by analogy" is good and, indeed, plugging $Y=Y_1$ or $Y=Y_2$ in (*) gives an identity.
Edit: (I had overlooked that you asked for s.p.d. solutions)
Now that we have found solutions, are they s.p.d. ?
Only $Y_2$ is s.p.d. Indeed it is the only one with a positive spectrum:
If $X=PDP^{-1}$, then it suffices to write $Y$ under the form 
$$Y=\dfrac{1}{2}(PDP^{-1}+P\Delta^{1/2}P^{-1})=\dfrac{1}{2}P(D+\Delta^{1/2})P^{-1}=\cdots=PEP^{-1} \ \ (1)$$
with $E=diag(\dfrac{1}{2}(\lambda_k+\sqrt{\lambda_k^2+4}))$ to verify that all eigenvalues are positive
(which is not the case if we take a minus sign in front of the square root symbol).
Here is a Matlab program that illustrates the computation of $Y_2$:
R=rand(3);
X=R'*R;
Y=0.5*(X+sqrtm(X*X+4*eye(3)));
X
Y-inv(Y),% the same as X

A: Your equation is equivalent to $$Y^2-XY=I$$ or $$(Y-\frac12X)^2=I+\frac14X^2$$
note that $XY=YX$.
