1
JEE Main 2024 (Online) 9th April Morning Shift
MCQ (Single Correct Answer)
+4
-1

The solution of the differential equation $$(x^2+y^2) \mathrm{d} x-5 x y \mathrm{~d} y=0, y(1)=0$$, is :

A
$$\left|x^2-4 y^2\right|^5=x^2$$
B
$$\left|x^2-2 y^2\right|^6=x$$
C
$$\left|x^2-2 y^2\right|^5=x^2$$
D
$$\left|x^2-4 y^2\right|^6=x$$
2
JEE Main 2024 (Online) 9th April Morning Shift
MCQ (Single Correct Answer)
+4
-1

The solution curve, of the differential equation $$2 y \frac{\mathrm{d} y}{\mathrm{~d} x}+3=5 \frac{\mathrm{d} y}{\mathrm{~d} x}$$, passing through the point $$(0,1)$$ is a conic, whose vertex lies on the line :

A
$$2 x+3 y=-9$$
B
$$2 x+3 y=-6$$
C
$$2 x+3 y=9$$
D
$$2 x+3 y=6$$
3
JEE Main 2024 (Online) 8th April Evening Shift
MCQ (Single Correct Answer)
+4
-1

Let $$y=y(x)$$ be the solution curve of the differential equation $$\sec y \frac{\mathrm{d} y}{\mathrm{~d} x}+2 x \sin y=x^3 \cos y, y(1)=0$$. Then $$y(\sqrt{3})$$ is equal to:

A
$$\frac{\pi}{6}$$
B
$$\frac{\pi}{12}$$
C
$$\frac{\pi}{3}$$
D
$$\frac{\pi}{4}$$
4
JEE Main 2024 (Online) 8th April Morning Shift
MCQ (Single Correct Answer)
+4
-1

Let $$f(x)$$ be a positive function such that the area bounded by $$y=f(x), y=0$$ from $$x=0$$ to $$x=a>0$$ is $$e^{-a}+4 a^2+a-1$$. Then the differential equation, whose general solution is $$y=c_1 f(x)+c_2$$, where $$c_1$$ and $$c_2$$ are arbitrary constants, is

A
$$\left(8 e^x+1\right) \frac{d^2 y}{d x^2}-\frac{d y}{d x}=0$$
B
$$\left(8 e^x+1\right) \frac{d^2 y}{d x^2}+\frac{d y}{d x}=0$$
C
$$\left(8 e^x-1\right) \frac{d^2 y}{d x^2}-\frac{d y}{d x}=0$$
D
$$\left(8 e^x-1\right) \frac{d^2 y}{d x^2}+\frac{d y}{d x}=0$$
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