In Rayleigh - Ritz method we extremise the integral using an appropriate solution. Here we have to extremise

$$I = \int\limits_{0}^{1} F(x,y,y')dx$$

where $F(x,y,y') = xy + \frac{1}{2}y'^2..........................(1)$

Assume the trial solution of eq(1)

$$\bar{y}(x) = c_{0} + c_{1}x + c_{2}x^2$$

Given $y(0) = 0 \therefore c_{0} = 0$

Also $y(1) = c_{1} + c_{2} = 0 \therefore c_{2} = - c_{1} \\ \therefore \bar{y}(x) = c_{1}x - c_{1}x^2 = c_{1}x(1 - x) \\ \therefore y^{-1}(x) = c_{1} - 2c_{1}x = c_{1}(1 - 2x)$

Substituting the values of y(x) & y'(x) in integral we get,

$$\bar{I} = \int\limits_{0}^{1} (x[c_{!}x(1 - x)] + \frac{1}{2} [c_{1}(1 - 2x)]^2)dx \\ = c_{1} \int\limits_{0}^{1} (x^2 - x^3) + \frac{1}{2} (c_{1} (1 - 4x + 4x^2))dx \\ = c_{1} \bigg[\bigg(\frac{x^3}{3} - \frac{x^4}{4}\bigg) + \frac{1}{2} \bigg(c_{1}(x - 2x^2 + \frac{4}{3} x^2\bigg)\bigg)\bigg]_{0}^{1} \\ = c_{1}\bigg[\frac{1}{12} + c_{1}\frac{1}{6}\bigg] \\ = \frac{c_{1}}{12} + \frac{c_{1}^2}{6}$$

It's stationary values are given by

$$\frac{d\bar{I}}{dc_{1}} = 0 \hspace{0.5cm} \therefore \frac{1}{12} + \frac{1}{3}c_{1} = 0 \\ \therefore c_{1} = \frac{-1}{4}$$

Hence the approximate solution is

$$\bar{y}(x) = - \frac{1}{4}x(1 - x)$$