JEE Main 2023 (Online) 31st January Evening Shift | Differential Equations Question 45 | Mathematics

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

$\left(3 y^{2}-5 x^{2}\right) y \mathrm{~d} x+2 x\left(x^{2}-y^{2}\right) \mathrm{d} y=0$

such that $y(1)=1$. Then $\left|(y(2))^{3}-12 y(2)\right|$ is equal to :

$16 \sqrt{2}$

$32 \sqrt{2}$

JEE Main 2023 (Online) 31st January Morning Shift

MCQ (Single Correct Answer)

-1

Out of Syllabus

Let a differentiable function $$f$$ satisfy $$f(x)+\int_\limits{3}^{x} \frac{f(t)}{t} d t=\sqrt{x+1}, x \geq 3$$. Then $$12 f(8)$$ is equal to :

JEE Main 2023 (Online) 30th January Evening Shift

MCQ (Single Correct Answer)

-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 :

$\log _e|x+y|+\frac{x y}{(x+y)^2}=0$

$\log _e|x+y|-\frac{x y}{(x+y)^2}=0$

$\log _e|x+y|+\frac{2 x y}{(x+y)^2}=0$

$\log _e|x+y|-\frac{2 x y}{(x+y)^2}=0$

JEE Main 2023 (Online) 30th January Morning Shift

MCQ (Single Correct Answer)

-1

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

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

$$\exp \left(\frac{1-\pi}{4 \sqrt{2}}\right)$$

$$\exp \left(\frac{4-\pi}{4 \sqrt{2}}\right)$$

$$\exp \left(\frac{4+\pi}{4 \sqrt{2}}\right)$$

$$\exp \left(\frac{\pi-4}{4 \sqrt{2}}\right)$$

PREVIOUS NEXT

Questions Asked from Differential Equations (MCQ (Single Correct Answer))

Number in Brackets after Paper Indicates No. of Questions