Embibe Experts Solutions for Chapter: Differential Equations, Exercise 1: JEE Main - 24 June 2022 Shift 2

Let $y=y\left(x\right)$  be the solution of the differential equation $x{\log }_{e}x\frac{dy}{dx}+y=x^2{\log }_{e}x,(x>1)$. If $y\left(2\right)=2$, then $y\left(e\right)$ is equal to

Let the solution curve $y=y\left(x\right)$ of the differential equation $\frac{dy}{dx}-\frac{3x^5{\tan }^{-1}\left(x^3\right)}{{\left(1+x^6\right)}^{\frac{3}{2}}}y=2x$ $\exp \frac{x^3-{\tan }^{-1}{x}^3}{\sqrt{(1+x{)}^6}}$ pass through the origin. Then $y\left(1\right)$ is equal to:

The solution of the differential equation $\frac{dy}{dx}=-\left(\frac{x^2+3y^2}{3x^2+y^2}\right),y\left(1\right)=0$ is

Let a differentiable function $f$ satisfy $f\left(x\right)+{\int }_3^x\frac{f\left(t\right)}{t}dt=\sqrt{x+1},x\ge 3$. Then $12f\left(8\right)$ is equal to:

Let $y=y\left(x\right)$ be the solution of the differential equation $\left(3y^2-5x^2\right)ydx+2x\left(x^2-y^2\right)dy=0$ such that $y\left(1\right)=1$. Then $\left|\left(y\right(2){)}^3-12y(2)\right|$ is equal to :

If $y=y\left(x\right)$ is the solution curve of the differential equation $\frac{dy}{dx}+y\tan x=x\sec x, 0\le x\le \frac{\pi }{3}, y\left(0\right)=1$, then $y\left(\frac{\pi }{6}\right)$ is equal to

Let $f:\mathrm{ℝ}\to \mathrm{ℝ}$ be a differentiable function such that $f^{\text{'}}\left(x\right)+f\left(x\right)={\int }_0^{2}f\left(t\right)dt$. If $f\left(0\right)=e^{-2}$, then $2f\left(0\right)-f\left(2\right)$ is equal to _____ .

Let $\alpha x=\exp \left(x^{\beta }{y}^{\gamma }\right)$ be the solution of the differential equation $2x^{2}ydy-\left(1-x{y}^2\right)dx=0$, $x>0,y\left(2\right)=\sqrt{{\log }_{\mathrm{e}}2}$. Then $\alpha +\beta -\gamma$ equals :