IIT-JEE 1999 | Differential Equations Question 34 | Mathematics

IIT-JEE 1999

MCQ (Single Correct Answer)

-0.5

A solution of the differential equation
$${\left( {{{dy} \over {dx}}} \right)^2} - x{{dy} \over {dx}} + y = 0$$ is

$$y=2$$

$$y=2x$$

$$y=2x-4$$

$$y = 2{x^2} - 4$$

IIT-JEE 1998

MCQ (Single Correct Answer)

-0.5

The order of the differential equation whose general solution is given by
$$y = \left( {{C_1} + {C_2}} \right)\cos \left( {x + {C_3}} \right) - {C_4}{e^{x + {C_5}}},$$ where
$${C_1},{C_2},{C_3},{C_4},{C_5},$$ are arbitrary constants, is

$$5$$

$$4$$

$$3$$

$$2$$

JEE Advanced Subjects