Consider two solutions $$\,x\left( t \right)\,\,\,\, = \,\,\,{x_1}\left( t \right)\,\,$$ and $$x\left( t \right)\,\,\,\, = \,\,\,{x_2}\left( t \right)\,\,$$ of the differential equation
$$\,\,{{{d^2}x\left( t \right)} \over {d{t^2}}} + x\left( t \right) = 0,t > 0,\,\,$$ such that
$$\,{x_1}\left( 0 \right) = 1,{\left. {{{d{x_1}\left( t \right)} \over {dt}}} \right|_{t = 0}} = 0,$$ $$\,\,\,\,{x_2}\left( 0 \right) = 0,{\left. {{{d{x_2}\left( t \right)} \over {dt}}} \right|_{t = 0}} = 1$$

The wronskian $$\,w\left( t \right) = \left| {{{\matrix{ {{x_1}\left( t \right)} \cr {d{x_1}\left( t \right)} \cr } } \over {dt}}} \right.\left. {{{\matrix{ {{x_2}\left( t \right)} \cr {d{x_2}\left( t \right)} \cr } } \over {dt}}} \right|$$ at $$\,\,t = \pi /2$$

The definite integral $$\,\int\limits_1^3 {{1 \over x}} dx\,\,$$ is evaluated using Trapezoidal rule with a step size of $$1.$$ The correct answer is

The real root of the equation $$5x-2cosx=0$$ (up to two decimal accuracy) is

An analytic function of a complex variable $$z = x + iy$$ is expressed as
$$f\left( z \right) = u\left( {x + y} \right) + iv\left( {x,y} \right),$$ where $$i = \sqrt { - 1} .$$ If $$u(x, y)=$$ $${x^3} - {y^2}$$
then expression for $$v(x,y)$$ in terms of $$x,y$$ and a general constant $$c$$ would be