written 5.0 years ago by | • modified 5.0 years ago |
Solution:
$I=\int_0^1 (xy +\frac {1}{2}y'^2)dx$
where $F=xy+ \frac {1}{2}y'^2$-----(2)
Assume the trial solution ,
$\bar y(x)=c_0+c_1x+c_2 x^2$ ----(3)
put $x=0$ in eqn(3)
$\bar y(0)=c_0+c_1(0)+c_2 (0)$
But $\bar y(0)=0$
$0=c_0+(0)+(0)$
$0=c_0$
Put $x=1$ in eqn(3)
$y_(1) = c_0+c_1(1)+c_2(1)$
but $\bar y (1) =0 $ and $c_0 =0$
$0 =0+c_1+c_2$
$c_2=-c_1$
Put $c_0=0$ and $c_2 =-c_1$ in eqn (3)
$\bar y (x) = 0+c_1x-c_1x^2$
$\bar y (x) = c_1(x-x^2)----(4)$
diff.(4) w.r.t x
$\bar y'(x)=c_1(1-2x)----(5)$
Put these values in eqn(1)
$I=\int_0^1 \{x[c_1(x-x^2)]+\frac{1}{2}c_1^2(1-2x)^2\} dx$
$I=c_1 \int_0^1 \{(x^2-x^3)]+\frac{1}{2}c_1(1-4x+4x^2)\} dx$
$c_1\{[\frac{x^3}{3} -\frac{x^4}{4}]+\frac{1}{2} c_1[1-\frac{4x^2}{2}+\frac{4x^3}{3}]\}_0^1 $
upper limit -lower limit
$c_1\{[\frac{1}{3} -\frac{1}{4}]+\frac{1}{2} c_1[1-\frac{4}{2}+\frac{4}{3}]\} -c_1\{[0-0]-\frac{1}{2} c_1[0-0+0]\}$
= $c_1\{\frac{1}{12} +\frac{1}{2} c_1 (\frac{1}{3})\}-c_1[0]$
= $c_1\{\frac{1}{12} +\frac{1}{2} c_1 \frac{1}{3)}\}$
$I=c_1(\frac{1}{12}) +c_1^2(\frac{1}{6})------(6) $
diff eqn (6) w.r.t $c_1$
$\frac {dI}{dc_1} =\frac{1}{12} +2c_1(\frac{1}{6})$
but $\frac {dI}{dc_1}$ is stationary value ,
hence $\frac{dI}{dc_1}=0$
$0= \frac{1}{12} +c_1(\frac{1}{3}) $
$\frac{-1}{12} = c_1(\frac{1}{3}) $
$c_1 = \frac{-1}{4}$
put $c_1 = \frac{-1}{4}$ in eqn (4)
$\bar y(x) = -\frac{1}{4} (x-x^2)$
$\bar y(x) = -\frac{1}{4} x(1-x)$