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

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

where $F(x,y,y') = 2xy - y^2 - y'^2........................(1)$

Assume the trial solution of eq(1)

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

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

Aksi y(1) = $c_{1} + c_{2} = 0 \hspace{0.25cm} \therefore c_{2} = -c_{1} \\ \therefore \bar{y}(x) = c_{1}x - c_{1}x^2 = c_{1}x (1 - x) \\ \therefore y^{-1} (x) = c_{1}x - 2c_{1}x = c_{1}(1 - 2x)$

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

$$\bar{I} = \int\limits_{0}^{1} (2x [ c_{1}x (1 - x)] - [c_{1}x (1 - x)]^2 - [c_{1}(1 - 2x)]^2)dx \\ = c_{1} \int\limits_{0}^{1} [2 (x^2 - x^3) - c_{1} (x^2 - 2x^3 + x^4 - (1 - 4x + 4x^2))]dx \\ = c_{1} \int\limits_{0}^{1}[2(x^2 - x^3) - c_{1}(-1 + 4x - 3x^2 - 2x^3 + x^4))]dx \\ = c_{1}\bigg[2 \bigg(\frac{x^3}{3} - \frac{x^4}{4} \bigg) - c_{1} \bigg(-x + 2x^2 - x^3 - \frac{x^4}{2} + \frac{x^5}{5} \bigg) \bigg]_{0}^{1} \\ = c_{1} \bigg[\frac{1}{6} - \frac{3}{10}c_{1}\bigg] = \frac{c_{1}}{6} - \frac{3}{10}c_{1}^2$$

It's stationary values are given by

$$\frac{d\bar{I}}{dc_{1}} = 0 \hspace{0.5cm} \therefore \frac{1}{6} - \frac{3}{5}c_{1} = 0 \\ \therefore c_{1} = \frac{5}{18}$$

Hence the approximate solution is

$$\bar{y}(x) = \frac{5}{18}x ( 1 - x)$$