Solution

$I=\int_0^1 F(x,y ,y')dx$-----(1)

where $F=2xy+y^2-y'^2$------(2)

Now assume the trial solution,

$\bar y(x)=c_0+c_1x+c_2x^2$---(3)

Put x=0 in (3)

$\bar y(0)=c_0+c_1(0)+c_2(0^2)$

$\bar y(0)=c_0+0+0$

But $\bar y(0)=0$--------given

$c_0=0$

Put x=1 in (3)

$\bar y(1) = c_0+c_1(1)+c_2(1)^2$

$\bar y(1) = 0+c_1+c_2$

$\bar y(1) = 0$----------(Given)

$0=c_1+c_2$

$c_2=-c_1$

Eqn (3) becomes

$\bar y(x)=0+c_1x+c_2x^2$

$\bar y(x)=c_1-c_1x^2$

$\bar y(x)=c_1x(1-x)$----(4)

diff. w r t x

$\bar y'(x)=c_1(1-2x)$-----(5)

Put all these values in equation (1)

$I=\int_0^1 F(2xy+y^2-y'2)dx$

$I=\int_0^1 [2xc_1x(1-x)+c_1^2 x^2 (1-x)^2-c_1^2(1-2_x)^2]dx$

$I=\int_0^1 [2c_1(x^2-x^3)+c_1^2 x^2 (1-2x+x^2)-c_1^2(1-4x+4x^2)]dx$

$I=\int_0^1 [2c_1(x^2-x^3)+c_1^2 (x^2-2x^3+x^4)-c_1^2(1-4x+4x^2)]dx$

$I=c_1\int_0^1 [2(x^2-x^3)+c_1[ x^2 +x^4-2x^3-1+4x-4x^2]dx$

$I=c_1\int_0^1 [2(x^2-x^3)+c_1[ -1+4x-3x^2-2x^3+x^4]dx$

$I=c_1 [2(\frac{x^3}{3}-\frac{x^4}{4})+c_1[ -x +\frac{4x^2}{2}-\frac{3x^3}{3}-\frac{2x^4}{4}-\frac{x^5}{5}]]^1_0$

upperlimit-lowerimit

$c_1{[2(\frac{1}{3}-\frac{1}{4})+c_1(-1+\frac{4}{2}-\frac{3}{3}-\frac{2}{4}+\frac{1}{5})]-[0]}$

$c_1 \{2[\frac{1}{12}]+c_1 [\frac{-3}{10}] \}$

$c_1[\frac{1}{6}-\frac {3}{10}c_1]$

$I=\frac {c_1}{6}-\frac{3}{10}c_1^2$----(6)

It is stationary values

$\frac {dI}{dc_1}=0$-------(7)

diff.(6)w.r.t $c_1$

$\frac{dI}{dc_1} =\frac {1}{6}-\frac{3}{10}2c_1$

$0=\frac{1}{6}-\frac{3}{5}c_1$----------(from 7)

$\frac{1}{6}=\frac{3}{5}c_1$

$c_1=\frac{5}{18}$

Put $c_1$ value in equation (4)

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