Solve the Ode: $$ \begin{cases} \frac{d^2y}{dx^2}+(\frac{1}{y})\frac{dy}{dx}=-1 \quad \hbox{for $0<y<1$} \\ y(1)=y'(0)=0. \end{cases} $$
Putting $\frac{dy}{dx}=p$ then $\frac{d^2y}{dx^2}=p\frac{dp}{dy}$ so the equation reduces to $\frac{dp}{dy}+\frac{1}{y}=-\frac{1}{p}$.
Then how to proceed.
On
Where is no closed form for the solutions of the non linear ODE : $$ \frac{d^2y}{dx^2}+(\frac{1}{y})\frac{dy}{dx}=-1 $$ Of course, it can be solved thanks to numerical methods, but it is not what is expected.
In your additional comment, you wrote :
" Answer is $(1−x^2 )/4$ .....but i dont know how to solve ..."
It is easy to check that $(1−x^2 )/4$ is not a solution of the above ODE.
Most likely there is a typo in the writing of the ODE, which might be : $$ \begin{cases} \frac{d^2y}{dx^2}+(\frac{1}{x})\frac{dy}{dx}=-1 \quad \hbox{for $0<y<1$} \\ y(1)=y'(0)=0. \end{cases} $$ This is a linear ODE easy to solve : $$\frac{dy}{dx}=\frac{c_1}{x}-\frac{x}{2}$$ $$y=c_1 \ln(x)-\frac{x^2}{4}+c_2$$ and the conditions $y(1)=y'(0)=0$ imply $c_1=0$ and $c_2$=1 $$y=1-\frac{x^2}{4}$$
$$ y''+\frac{1}{y}y' +1 = 0 $$ this becomes $$ \dfrac{d}{dx}\left(y' + \ln y + x\right) = 0 $$ thus $$ y' + \ln y + x = C_1 $$ but with you initial conditions it is clear that the solution is not bounded. take $y(1) = 0$ we get $$ y'(1) - \infty + 1 = C_1\\ 0 +\ln y(0) =C_1. $$ which is sort of silly to me. Besides the solution to Eq.(*) Myself (and Wolfram) failed to find a solution.