How to solve this differential equation efficiently?
$$\begin{align} 2yy'' &= y'^2 + y^2, \\ y(0) &= 1,\\y'(0) &= -1. \end{align}$$
My method includes setting $p=y'$, $t=\dfrac{p}{y}$,$u=t^2$, which turns the original equation into linear equation $$ \frac{\text{d}u}{\text{d}y}+\frac1y u=\frac1y. $$
It requires so much substitutions.
Are there any other methods?
Hint
For the time being, do not consider the conditions and switch variables. The differential equation becomes $$2y\,x''(y)+x'(y) \left(y^2\, x'(y)^2+1\right)=0.$$ Reduction of order $p(y)=x'(y)$ leads to $$p(y)=\pm \frac 1 {\sqrt{y(y+c_1) }}=x'(y),$$ which is simple to integrate and to inverse.
Now, use the conditions (hoping that they are consistent).
Edit
Staying with the original equation, use the ansatz $$y(x)=z(x)\, e^{a x^b}$$ and search for $(a,b)$ which eliminate most of the terms. You should end with a very simple differential equation.
We have
$$ 2\frac{y''}{y}=\left(\frac{y'}{y}\right)^2+1 $$
but
$$ \left(\frac{y'}{y}\right)'=\frac{y''}{y}-\left(\frac{y'}{y}\right)^2 $$
hence
$$ 2\left(\frac{y'}{y}\right)^2+2\left(\frac{y'}{y}\right)'=\left(\frac{y'}{y}\right)^2+1 $$
or
$$ 2\left(\frac{y'}{y}\right)'+\left(\frac{y'}{y}\right)^2-1=0 $$
Solving now
$$ 2u'+u^2-1=0\Rightarrow u = \frac{e^t-c_1}{e^t+c_1} $$
and finally
$$ y' = \frac{e^t-c_1}{e^t+c_1}y\Rightarrow y = c_2 e^{-t} \left(e^t+c_1\right)^2 $$
If you wish to avoid substitutions you can integrate directly. Multiply by $y'/y^2$ to get an exact equation (when integrating I denote $y':=\mathrm dy(\xi)/\mathrm d\xi$ where $\xi$ is a dummy variable): \begin{align} \int_0^x\left( \frac{2y'}{y}y''-\frac{y'^3}{y^2}-y'\right) \mathrm d\xi=0,\\\\ \frac{y'^2}{y}-\frac{(-1)^2}{1}-y+1=\frac{y'^2}{y}-y=0, \end{align} and so \begin{align} y'&=-y,\\\\ \int_0^x\frac{y'}{y}\mathrm d\xi&=-\int_0^x \mathrm d\xi,\\\\ \log y-\log 1=- (x-0)&\implies y=e^{- x}. \end{align}
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.
Another method using substitution (which I do not think I've seen mentioned above).
Since this ode is missing $x$, then let \begin{align*} \frac{dy}{dx}=p \end{align*} And now \begin{align*} y''&=\frac{dp}{dx}\\ &=\frac{dp}{dy} \frac{dy}{dx}\\ &=p' p \end{align*}
The original ode becomes \begin{align*} 2 y y'' &= (y')^2 + y^2\\ 2y p' p &=p^2 + y^2\\ p' &= \frac{p}{2y}+ \frac{y}{2p} \\ p' - \frac{p}{2y} &= \frac{y}{2p} \end{align*}
Which has the form $ p' + Q(y) p = R(y) p^n $ where here $Q(y)=-\frac{1}{2y}$ and $R(y)=\frac{y}{2}$ and $n=-1$. Hence it is standard first order Bernoulli ode which can be solved for $p$ using standard Bernoulli ode algorithm.
Once $p$ is known (which will be a function of $y$), then $y$ is solved from the first order ode $\frac{dy}{dx}=p$
So this method requires solving two first order ode's.
