A boundary value problem for the ODE $y''+y'/y=-1$ 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.
 A: $$
y''+\frac{1}{y}y' +1 = 0
$$
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.
A: 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
$$y=1-\frac{x^2}{4}$$
