1
JEE Main 2023 (Online) 29th January Evening Shift
+4
-1

Let $$y=y(x)$$ be the solution of the differential equation $$x{\log _e}x{{dy} \over {dx}} + y = {x^2}{\log _e}x,(x > 1)$$. If $$y(2) = 2$$, then $$y(e)$$ is equal to

A
$${{1 + {e^2}} \over 2}$$
B
$${{1 + {e^2}} \over 4}$$
C
$${{2 + {e^2}} \over 2}$$
D
$${{4 + {e^2}} \over 4}$$
2
JEE Main 2023 (Online) 29th January Morning Shift
+4
-1

Let $$y=f(x)$$ be the solution of the differential equation $$y(x+1)dx-x^2dy=0,y(1)=e$$. Then $$\mathop {\lim }\limits_{x \to {0^ + }} f(x)$$ is equal to

A
$${e^2}$$
B
0
C
$${1 \over {{e^2}}}$$
D
$${1 \over e}$$
3
JEE Main 2023 (Online) 25th January Evening Shift
+4
-1

Let $$y=y(t)$$ be a solution of the differential equation $${{dy} \over {dt}} + \alpha y = \gamma {e^{ - \beta t}}$$ where, $$\alpha > 0,\beta > 0$$ and $$\gamma > 0$$. Then $$\mathop {\lim }\limits_{t \to \infty } y(t)$$

A
is 0
B
is 1
C
is $$-1$$
D
does not exist
4
JEE Main 2023 (Online) 25th January Morning Shift
+4
-1

Let $$y = y(x)$$ be the solution curve of the differential equation $${{dy} \over {dx}} = {y \over x}\left( {1 + x{y^2}(1 + {{\log }_e}x)} \right),x > 0,y(1) = 3$$. Then $${{{y^2}(x)} \over 9}$$ is equal to :

A
$${{{x^2}} \over {5 - 2{x^3}(2 + {{\log }_e}{x^3})}}$$
B
$${{{x^2}} \over {3{x^3}(1 + {{\log }_e}{x^2}) - 2}}$$
C
$${{{x^2}} \over {7 - 3{x^3}(2 + {{\log }_e}{x^2})}}$$
D
$${{{x^2}} \over {2{x^3}(2 + {{\log }_e}{x^3}) - 3}}$$
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