For this problem alone, there is an interesting short-cut. Try to guess what $y$ might be from the given condition. We know that, $y(0) = 1$ while $y'(0) = -1$. This means whatever the function is, the derivative of it yields the same value with the negative sign. The function that has the same value as its derivative is $e^x$. So, this function $y$ should be $e^{-x}$.

Do a little sanity check.

$y(0) = e^{-0} = 1$, $y'(0) = -e^{-0} = -1$, and $y'' = e^{-x}$

Substitute it in your equation,

$2yy'' = y'^2 + y^2$

$2e^{-x}e^{-x} = (-e^{-x})^2 + (e^{-x})^2$

$2e^{-2x} = e^{-2x} + e^{-2x} = 2e^{-2x}$

While you may need to add some constant C to the function since you would be integrating it. But there you go. Probably the easiest way.

For this problem alone, there is an interesting short-cut. Try to guess what $y$ might be from the given condition. We know that, $y(0) = 1$ while $y'(0) = -1$. This means whatever the function is, the derivative of it yields the same value with the negative sign. The function that has the same value as its derivative is $e^x$. So, this function $y$ should be $e^{-x}$.

Do a little sanity check.

$y(0) = e^{-0} = 1$, $y'(0) = -e^{-0} = -1$, and $y'' = e^{-x}$

Substitute it in your equation,

$2yy'' = y'^2 + y^2$

$2e^{-x}e^{-x} = (-e^{-x})^2 + (e^{-x})^2$

$2e^{-2x} = e^{-2x} + e^{-2x} = 2e^{-2x}$

While you may need to add some constant $C$ to the function since you would be integrating it. But there you go. Probably the easiest way.