JEE Main 2024 (Online) 6th April Morning Shift | Differential Equations Question 11 | Mathematics

Let $$y=y(x)$$ be the solution of the differential equation $$\left(2 x \log _e x\right) \frac{d y}{d x}+2 y=\frac{3}{x} \log _e x, x>0$$ and $$y\left(e^{-1}\right)=0$$. Then, $$y(e)$$ is equal to

$$-\frac{3}{\mathrm{e}}$$

$$-\frac{3}{2 \mathrm{e}}$$

$$-\frac{2}{3 \mathrm{e}}$$

$$-\frac{2}{\mathrm{e}}$$

JEE Main 2024 (Online) 6th April Morning Shift

MCQ (Single Correct Answer)

-1

Let $$y=y(x)$$ be the solution of the differential equation $$\left(1+x^2\right) \frac{d y}{d x}+y=e^{\tan ^{-1} x}$$, $$y(1)=0$$. Then $$y(0)$$ is

$$\frac{1}{4}\left(e^{\pi / 2}-1\right)$$

$$\frac{1}{2}\left(1-e^{\pi / 2}\right)$$

$$\frac{1}{4}\left(1-e^{\pi / 2}\right)$$

$$\frac{1}{2}\left(e^{\pi / 2}-1\right)$$

JEE Main 2024 (Online) 5th April Evening Shift

MCQ (Single Correct Answer)

-1

The differential equation of the family of circles passing through the origin and having centre at the line $$y=x$$ is :

$$\left(x^2-y^2+2 x y\right) \mathrm{d} x=\left(x^2-y^2+2 x y\right) \mathrm{d} y$$

$$\left(x^2+y^2-2 x y\right) \mathrm{d} x=\left(x^2+y^2+2 x y\right) \mathrm{d} y$$

$$\left(x^2+y^2+2 x y\right) \mathrm{d} x=\left(x^2+y^2-2 x y\right) \mathrm{d} y$$

$$\left(x^2-y^2+2 x y\right) \mathrm{d} x=\left(x^2-y^2-2 x y\right) \mathrm{d} y$$

JEE Main 2024 (Online) 5th April Morning Shift

MCQ (Single Correct Answer)

-1

If $$y=y(x)$$ is the solution of the differential equation $$\frac{\mathrm{d} y}{\mathrm{~d} x}+2 y=\sin (2 x), y(0)=\frac{3}{4}$$, then $$y\left(\frac{\pi}{8}\right)$$ is equal to :

$$\mathrm{e}^{-\pi / 8}$$

$$\mathrm{e}^{\pi / 4}$$

$$\mathrm{e}^{-\pi / 4}$$

$$\mathrm{e}^{\pi / 8}$$

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