1
JEE Main 2024 (Online) 31st January Morning Shift
+4
-1

The solution curve of the differential equation $$y \frac{d x}{d y}=x\left(\log _e x-\log _e y+1\right), x>0, y>0$$ passing through the point $$(e, 1)$$ is

A
$$\left|\log _e \frac{y}{x}\right|=y^2$$
B
$$\left|\log _e \frac{y}{x}\right|=x$$
C
$$\left|\log _e \frac{x}{y}\right|=y$$
D
$$2\left|\log _e \frac{x}{y}\right|=y+1$$
2
JEE Main 2024 (Online) 30th January Morning Shift
+4
-1

Let $$y=y(x)$$ be the solution of the differential equation $$\sec x \mathrm{~d} y+\{2(1-x) \tan x+x(2-x)\} \mathrm{d} x=0$$ such that $$y(0)=2$$. Then $$y(2)$$ is equal to:

A
$$2\{\sin (2)+1\}$$
B
2
C
1
D
$$2\{1-\sin (2)\}$$
3
JEE Main 2024 (Online) 29th January Evening Shift
+4
-1

If $$\sin \left(\frac{y}{x}\right)=\log _e|x|+\frac{\alpha}{2}$$ is the solution of the differential equation $$x \cos \left(\frac{y}{x}\right) \frac{d y}{d x}=y \cos \left(\frac{y}{x}\right)+x$$ and $$y(1)=\frac{\pi}{3}$$, then $$\alpha^2$$ is equal to

A
12
B
9
C
4
D
3
4
JEE Main 2024 (Online) 29th January Morning Shift
+4
-1

A function $$y=f(x)$$ satisfies $$f(x) \sin 2 x+\sin x-\left(1+\cos ^2 x\right) f^{\prime}(x)=0$$ with condition $$f(0)=0$$. Then, $$f\left(\frac{\pi}{2}\right)$$ is equal to

A
2
B
1
C
$$-$$1
D
0
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