1
JEE Main 2023 (Online) 30th January Evening Shift
MCQ (Single Correct Answer)
+4
-1
The solution of the differential equation

$\frac{d y}{d x}=-\left(\frac{x^2+3 y^2}{3 x^2+y^2}\right), y(1)=0$ is :
A
$\log _e|x+y|+\frac{x y}{(x+y)^2}=0$
B
$\log _e|x+y|-\frac{x y}{(x+y)^2}=0$
C
$\log _e|x+y|+\frac{2 x y}{(x+y)^2}=0$
D
$\log _e|x+y|-\frac{2 x y}{(x+y)^2}=0$
2
JEE Main 2023 (Online) 30th January Morning Shift
MCQ (Single Correct Answer)
+4
-1

Let the solution curve $$y=y(x)$$ of the differential equation

$$\frac{\mathrm{d} y}{\mathrm{~d} x}-\frac{3 x^{5} \tan ^{-1}\left(x^{3}\right)}{\left(1+x^{6}\right)^{3 / 2}} y=2 x \exp \left\{\frac{x^{3}-\tan ^{-1} x^{3}}{\sqrt{\left(1+x^{6}\right)}}\right\} \text { pass through the origin. Then } y(1) \text { is equal to : }$$

A
$$\exp \left(\frac{1-\pi}{4 \sqrt{2}}\right)$$
B
$$\exp \left(\frac{4-\pi}{4 \sqrt{2}}\right)$$
C
$$\exp \left(\frac{4+\pi}{4 \sqrt{2}}\right)$$
D
$$\exp \left(\frac{\pi-4}{4 \sqrt{2}}\right)$$
3
JEE Main 2023 (Online) 29th January Evening Shift
MCQ (Single Correct Answer)
+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
$${{4 + {e^2}} \over 4}$$
4
JEE Main 2023 (Online) 29th January Morning Shift
MCQ (Single Correct Answer)
+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
$${1 \over e}$$
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