Value of the integral $$\,\,\oint {xydy - {y^2}dx,\,\,} $$ where, $$c$$ is the square cut from the first quadrant by the line $$x=1$$ and $$y=1$$ will be (Use Green's theorem to change the line integral into double integral)

The solution $${{{d^2}y} \over {d{x^2}}} + 2{{dy} \over {dx}} + 17y = 0;$$ $$y\left( 0 \right) = 1,{\left( {{{d\,y} \over {d\,x}}} \right)_{x = {\raise0.5ex\hbox{$\scriptstyle \pi $} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 4$}}}} = 0\,\,$$ in the range $$0 < x < {\pi \over 4}$$ is given by

Transformation to linear form by substituting $$v = {y^{1 - n}}$$ of the equation $${{dy} \over {dt}} + p\left( t \right)y = q\left( t \right){y^n},\,\,n > 0$$ will be

Given $$a>0,$$ we wish to calculate its reciprocal value $${1 \over a}$$ by using Newton - Raphson method for $$f(x)=0.$$ For $$a=7$$ and starting with $${x_0} = 0.2\,\,$$ the first two iterations will be