1
JEE Main 2019 (Online) 9th April Morning Slot
+4
-1
The solution of the differential equation

$$x{{dy} \over {dx}} + 2y$$ = x2 (x $$\ne$$ 0) with y(1) = 1, is :
A
$$y = {4 \over 5}{x^3} + {1 \over {5{x^2}}}$$
B
$$y = {3 \over 4}{x^2} + {1 \over {4{x^2}}}$$
C
$$y = {{{x^2}} \over 4} + {3 \over {4{x^2}}}$$
D
$$y = {{{x^3}} \over 5} + {1 \over {5{x^2}}}$$
2
JEE Main 2019 (Online) 8th April Morning Slot
+4
-1
Let y = y(x) be the solution of the differential equation,

$${({x^2} + 1)^2}{{dy} \over {dx}} + 2x({x^2} + 1)y = 1$$

such that y(0) = 0. If $$\sqrt ay(1)$$ = $$\pi \over 32$$ , then the value of 'a' is :
A
$${1 \over 2}$$
B
$${1 \over 16}$$
C
1
D
$${1 \over 4}$$
3
JEE Main 2019 (Online) 12th January Evening Slot
+4
-1
If a curve passes through the point (1, –2) and has slope of the tangent at any point (x, y) on it as $${{{x^2} - 2y} \over x}$$, then the curve also passes through the point :
A
(–1, 2)
B
$$\left( { - \sqrt 2 ,1} \right)$$
C
$$\left( { \sqrt 3 ,0} \right)$$
D
(3, 0)
4
JEE Main 2019 (Online) 12th January Morning Slot
+4
-1
Let y = y(x) be the solution of the differential equation, x$${{dy} \over {dx}}$$ + y = x loge x, (x > 1). If 2y(2) = loge 4 $$-$$ 1, then y(e) is equal to :
A
$$- {e \over 2}$$
B
$$- {{{e^2}} \over 2}$$
C
$${{{e^2}} \over 4}$$
D
$${e \over 4}$$
EXAM MAP
Medical
NEET