JEE Advance - Mathematics (1999 - No. 4)
A solution of the differential equation
$${\left( {{{dy} \over {dx}}} \right)^2} - x{{dy} \over {dx}} + y = 0$$ is
$${\left( {{{dy} \over {dx}}} \right)^2} - x{{dy} \over {dx}} + y = 0$$ is
$$y=2$$
$$y=2x$$
$$y=2x-4$$
$$y = 2{x^2} - 4$$
Comments (0)
