If any differential equation is given by $f''(x)+f'(x)+f^2(x) = x^2\;,$ Then $f(x)=$ 
If any differential equation is given by $f''(x)+f'(x)+f^2(x) = x^2\;,$ Then $f(x)=$ 

$\bf{My\; Try:}$ 
We can write the above differential  equation as $$e^xf''(x)+e^xf'(x)+e^x\cdot (f(x))^2 = e^x\cdot x^2$$
So $$\frac{d}{dx}\left[e^x\cdot f'(x)\right] = e^x\cdot x^2-e^x(f(x))^2$$
Now How can I proceed after that, help me
Thanks.

Using the power series trick we get that:
\begin{align}
 &f(x) = \sum_{k=0}^\infty a_kx^k\\
 &f'(x) = \sum_{k=0}^\infty a_{k+1}(k+1) x^k\\
 &f''(x) = \sum_{k=0}^\infty a_{k+2}(k+1)(k+2)x^k\\
 &f(x)^2 = \sum_{k=0}^\infty \left(\sum_{i=0}^k a_ia_{k-i} \right)x^k
\end{align}
Inserting this into the equation we have that:
$$
  (k+1)(k+2)a_{k+2}+(k+1)a_{k+1}+\sum_{i=0}^k a_ia_{k-i} = \delta_{k,2}, \quad \mbox{for } k\geq 0.
$$
Noting that $a_0 = f(0)$ we may successively solve the recurrence equation to find a solution. As a side remark we know (since the coefficients in the ODE are entire) that the radius of convergence is infinite (Fuchs theorem - if i recall correctly).
