If $$y = {x \over {{{\log }_e}|cx|}}$$ is the solution of the differential equation $${{dy} \over {dx}} = {y \over x} + \phi \left( {{x \over y}} \right)$$, then $$\phi \left( {{x \over y}} \right)$$ is given by
The function $$y = {e^{kx}}$$ satisfies $$\left( {{{{d^2}y} \over {d{x^2}}} + {{dy} \over {dx}}} \right)\left( {{{dy} \over {dx}} - y} \right) = y{{dy} \over {dx}}$$. It is valid for
Given $${{{d^2}y} \over {d{x^2}}} + \cot x{{dy} \over {dx}} + 4y\cos e{c^2}x = 0$$. Changing the independent variable x to z by the substitution $$z = \log \tan {x \over 2}$$, the equation is changed to
Let $$f(x) = \left\{ {\matrix{ {x + 1,} & { - 1 \le x \le 0} \cr { - x,} & {0 < x \le 1} \cr } } \right.$$
WB JEE Papers
All year-wise previous year question papers