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

$ x \frac{d y}{d x}-\sin 2 y=x^3\left(2-x^3\right) \cos ^2 y, x \neq 0 . $

If $y(2)=0$, then $\tan (y(1))$ is equal to

Let $y=y(x)$ be the solution of the differential equation $x^4 \mathrm{~d} y+\left(4 x^3 y+2 \sin x\right) \mathrm{d} x=0, x>0, y\left(\frac{\pi}{2}\right)=0$.

Then $\pi^4 y\left(\frac{\pi}{3}\right)$ is equal to :

If $y=y(x)$ satisfies the differential equation $16(\sqrt{x+9 \sqrt{x}})(4+\sqrt{9+\sqrt{x}}) \cos y \mathrm{~d} y=(1+2 \sin y) \mathrm{d} x, x>0$ and $y(256)=\frac{\pi}{2}, y(49)=\alpha$, then $2 \sin \alpha$ is equal to :

Let the solution curve of the differential equation $x d y-y d x=\sqrt{x^2+y^2} d x, x>0$, $y(1)=0$, be $y=y(x)$. Then $y(3)$ is equal to

Let $y = y(x)$ be the solution of the differential equation $\sec x \dfrac{dy}{dx} - 2y = 2 + 3 \sin x$, $x \in \left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$,

$y(0) = -\dfrac{7}{4}$. Then $y\left(\dfrac{\pi}{6}\right)$ is equal to :

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

Let $y=y(x)$ be the solution of the differential equation $x \sqrt{1-x^2} d y+\left(y \sqrt{1-x^2}-x \cos ^{-1} x\right) d x=0, x \in(0,1), \lim _{x \rightarrow 1^{-}} y(x)=1$. Then $y\left(\frac{1}{2}\right)$ equals :

Let $f: \mathbf{R} \rightarrow \mathbf{R}$ be such that $f(x y)=f(x) f(y)$, for all $x, y \in \mathbf{R}$ and $f(0) \neq 0$. Let $g:[1, \infty) \rightarrow \mathbf{R}$ be a differentiable function such that

$ x^2 g(x)=\int_1^x\left(\mathrm{t}^2 f(\mathrm{t})-\operatorname{tg}(\mathrm{t})\right) d t $

Then $g(2)$ is equal to :

Let $f:[1, \infty) \rightarrow \mathbf{R}$ be a differentiable function defined as $f(x)=\int_1^x f(\mathrm{t}) \mathrm{dt}+(1-x)\left(\log _{\mathrm{e}} x-1\right)+\mathrm{e}$.

Then the value of $f(f(1))$ is :

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

$ \frac{d y}{d x}+\left(\frac{6 x^2+\left(3 x^2+2 x^3+4\right) e^{-2 x}}{\left(x^3+2\right)\left(2+e^{-2 x}\right)}\right) y=2+e^{-2 x} $

$x \in(-1,2)$, satisfying $y(0)=\frac{3}{2}$. If $y(1)=\alpha\left(2+e^{-2}\right)$, then $\alpha$ is equal to :

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

Let $x = x(y)$ be the solution of the differential equation $2y^2 \frac{dx}{dy} - 2xy + x^2 = 0$, $y > 1$, $x(e) = e$.

Then $x(e^2)$ is equal to:

If the curve $y = f(x)$ passes through the point $(1, e)$ and satisfies the differential equation $dy = y(2 + \\log_e x) dx$, $x > 0$, then $f(e)$ is equal to :

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

$(1 + \sin x)\dfrac{dy}{dx} + (y + 1)\cos x = 0,\ y(0) = 0.$ If the curve $y = y(x)$ passes through the point $\left( \alpha , \dfrac{-1}{2} \right)$,

then a value of $\alpha$ is:

Let $f(x) = x - 1$ and $g(x) = e^x$ for $x \in \mathbb{R}$. If $\frac{dy}{dx} = \left( e^{-2\sqrt{x}} g\left(f(f(x))\right) - \frac{y}{\sqrt{x}} \right)$, $y(0) = 0$, then $y(1)$ is

Let y = y(x) be the solution of the differential equation $(x^2 + 1)y' - 2xy = (x^4 + 2x^2 + 1)\cos x$,

$y(0) = 1$. Then $ \int\limits_{-3}^{3} y(x) \, dx $ is :

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

$x\left(x^2+e^x\right) d y+\left(\mathrm{e}^x(x-2) y-x^3\right) \mathrm{d} x=0, x>0$, passing through the point $(1,0)$. Then $y(2)$ is equal to :

If a curve $y=y(x)$ passes through the point $\left(1, \frac{\pi}{2}\right)$ and satisfies the differential equation $\left(7 x^4 \cot y-\mathrm{e}^x \operatorname{cosec} y\right) \frac{\mathrm{d} x}{\mathrm{~d} y}=x^5, x \geq 1$, then at $x=2$, the value of $\cos y$ is :

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

$\frac{d y}{d x}+3\left(\tan ^2 x\right) y+3 y=\sec ^2 x, y(0)=\frac{1}{3}+e^3$. Then $y\left(\frac{\pi}{4}\right)$ is equal to :

If for the solution curve $y=f(x)$ of the differential equation $\frac{d y}{d x}+(\tan x) y=\frac{2+\sec x}{(1+2 \sec x)^2}$, $x \in\left(\frac{-\pi}{2}, \frac{\pi}{2}\right), f\left(\frac{\pi}{3}\right)=\frac{\sqrt{3}}{10}$, then $f\left(\frac{\pi}{4}\right)$ is equal to:

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

$\cos x\left(\log _e(\cos x)\right)^2 d y+\left(\sin x-3 y \sin x \log _e(\cos x)\right) d x=0$, x ∈ (0, $\frac{\pi}{2}$ ). If $ y(\frac{\pi}{4}) $ = $-\frac{1}{\log_{e}2}$, then $ y(\frac{\pi}{6}) $ is equal to :

Let for some function $\mathrm{y}=f(x), \int_0^x t f(t) d t=x^2 f(x), x>0$ and $f(2)=3$. Then $f(6)$ is equal to

Let $\mathrm{y}=\mathrm{y}(\mathrm{x})$ be the solution of the differential equation $\left(x y-5 x^2 \sqrt{1+x^2}\right) d x+\left(1+x^2\right) d y=0, y(0)=0$. Then $y(\sqrt{3})$ is equal to

Let $x=x(y)$ be the solution of the differential equation $y=\left(x-y \frac{\mathrm{~d} x}{\mathrm{~d} y}\right) \sin \left(\frac{x}{y}\right), y>0$ and $x(1)=\frac{\pi}{2}$. Then $\cos (x(2))$ is equal to :

Let a curve $y=f(x)$ pass through the points $(0,5)$ and $\left(\log _e 2, k\right)$. If the curve satisfies the differential equation $2(3+y) e^{2 x} d x-\left(7+e^{2 x}\right) d y=0$, then $k$ is equal to

If $x=f(y)$ is the solution of the differential equation $\left(1+y^2\right)+\left(x-2 \mathrm{e}^{\tan ^{-1} y}\right) \frac{\mathrm{d} y}{\mathrm{~d} x}=0, y \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ with $f(0)=1$, then $f\left(\frac{1}{\sqrt{3}}\right)$ is equal to :

Let $x=x(y)$ be the solution of the differential equation $y^2 \mathrm{~d} x+\left(x-\frac{1}{y}\right) \mathrm{d} y=0$. If $x(1)=1$, then $x\left(\frac{1}{2}\right)$ is :

Let $f(x)$ be a real differentiable function such that $f(0)=1$ and $f(x+y)=f(x) f^{\prime}(y)+f^{\prime}(x) f(y)$ for all $x, y \in \mathbf{R}$. Then $\sum_\limits{n=1}^{100} \log _e f(n)$ is equal to :

Let $f: \mathbf{R} \rightarrow \mathbf{R}$ be a twice differentiable function such that $f(x+y)=f(x) f(y)$ for all $x, y \in \mathbf{R}$. If $f^{\prime}(0)=4 \mathrm{a}$ and $f$ satisfies $f^{\prime \prime}(x)-3 \mathrm{a} f^{\prime}(x)-f(x)=0, \mathrm{a}>0$, then the area of the region $\mathrm{R}=\{(x, y) \mid 0 \leq y \leq f(a x), 0 \leq x \leq 2\}$ is :

Let $\int_\limits0^x \sqrt{1-\left(y^{\prime}(t)\right)^2} d t=\int_0^x y(t) d t, 0 \leq x \leq 3, y \geq 0, y(0)=0$. Then at $x=2, y^{\prime \prime}+y+1$ is equal to

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

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 :

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:

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

Let $y=y(x)$ be the solution of the differential equation $(1+y^2) e^{\tan x} d x+\cos ^2 x(1+e^{2 \tan x}) d y=0, y(0)=1$. Then $y\left(\frac{\pi}{4}\right)$ is equal to

Suppose the solution of the differential equation $\frac{d y}{d x}=\frac{(2+\alpha) x-\beta y+2}{\beta x-2 \alpha y-(\beta \gamma-4 \alpha)}$ represents a circle passing through origin. Then the radius of this circle is :

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

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

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

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 :

Let $y=y(x)$ be the solution of the differential equation $(x^2+4)^2 d y+(2 x^3 y+8 x y-2) d x=0$. If $y(0)=0$, then $y(2)$ is equal to

If the solution $y=y(x)$ of the differential equation $(x^4+2 x^3+3 x^2+2 x+2) \mathrm{d} y-(2 x^2+2 x+3) \mathrm{d} x=0$ satisfies $y(-1)=-\frac{\pi}{4}$, then $y(0)$ is equal to :

The temperature $T(t)$ of a body at time $t=0$ is $160^{\circ} \mathrm{F}$ and it decreases continuously as per the differential equation $\frac{d T}{d t}=-K(T-80)$, where $K$ is a positive constant. If $T(15)=120^{\circ} \mathrm{F}$, then $T(45)$ is equal to

Let $y=y(x)$ be the solution of the differential equation $\frac{d y}{d x}=\frac{(\tan x)+y}{\sin x(\sec x-\sin x \tan x)}, x \in\left(0, \frac{\pi}{2}\right)$ satisfying the condition $y\left(\frac{\pi}{4}\right)=2$. Then, $y\left(\frac{\pi}{3}\right)$ is

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

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:

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 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

If $y=y(x)$ is the solution curve of the differential equation $\left(x^2-4\right) \mathrm{d} y-\left(y^2-3 y\right) \mathrm{d} x=0, x>2, y(4)=\frac{3}{2}$ and the slope of the curve is never zero, then the value of $y(10)$ equals :

Let $y=y_{1}(x)$ and $y=y_{2}(x)$ be the solution curves of the differential equation $\frac{d y}{d x}=y+7$ with initial conditions $y_{1}(0)=0$ and $y_{2}(0)=1$ respectively. Then the curves $y=y_{1}(x)$ and $y=y_{2}(x)$ intersect at

Let $y=y(x), y > 0$, be a solution curve of the differential equation $\left(1+x^{2}\right) \mathrm{d} y=y(x-y) \mathrm{d} x$. If $y(0)=1$ and $y(2 \sqrt{2})=\beta$, then

Let $y=y(x)$ be the solution of the differential equation $\frac{d y}{d x}+\frac{5}{x\left(x^{5}+1\right)} y=\frac{\left(x^{5}+1\right)^{2}}{x^{7}}, x > 0$. If $y(1)=2$, then $y(2)$ is equal to :

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

$\left(1-x^{2} y^{2}\right) d x=y d x+x d y$.

If the line $x=1$ intersects the curve $y=y(x)$ at $y=2$ and the line $x=2$ intersects the curve $y=y(x)$ at $y=\alpha$, then a value of $\alpha$ is :

Let $f$ be a differentiable function such that ${x^2}f(x) - x = 4\int\limits_0^x {tf(t)dt} $, $f(1) = {2 \over 3}$. Then $18f(3)$ is equal to :

If the solution curve $f(x, y)=0$ of the differential equation

$\left(1+\log _{e} x\right) \frac{d x}{d y}-x \log _{e} x=e^{y}, x > 0$,

passes through the points $(1,0)$ and $(\alpha, 2)$, then $\alpha^{\alpha}$ is equal to :

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

The area enclosed by the closed curve $\mathrm{C}$ given by the differential equation

$\frac{d y}{d x}+\frac{x+a}{y-2}=0, y(1)=0$ is $4 \pi$.

Let $P$ and $Q$ be the points of intersection of the curve $\mathrm{C}$ and the $y$-axis. If normals at $P$ and $Q$ on the curve $\mathrm{C}$ intersect $x$-axis at points $R$ and $S$ respectively, then the length of the line segment $R S$ is :

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

$\frac{d y}{d x}+y \tan x=x \sec x, 0 \leq x \leq \frac{\pi}{3}, y(0)=1$, then $y\left(\frac{\pi}{6}\right)$ is equal to

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 :

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 : } $

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

Let $y=f(x)$ be the solution of the differential equation $y(x+1)dx-x^2dy=0,y(1)=e$. Then $\mathop {\lim }\limits_{x \to {0^ + }} f(x)$ is equal to

Let $y=y(t)$ be a solution of the differential equation ${{dy} \over {dt}} + \alpha y = \gamma {e^{ - \beta t}}$ where, $\alpha > 0,\beta > 0$ and $\gamma > 0$. Then $\mathop {\lim }\limits_{t \to \infty } y(t)$

Let $y = y(x)$ be the solution curve of the differential equation ${{dy} \over {dx}} = {y \over x}\left( {1 + x{y^2}(1 + {{\log }_e}x)} \right),x > 0,y(1) = 3$. Then ${{{y^2}(x)} \over 9}$ is equal to :

Let $y=y(x)$ be the solution of the differential equation $(x^2-3y^2)dx+3xy~dy=0,y(1)=1$. Then $6y^2(e)$ is equal to

Let $y = y(x)$ be the solution of the differential equation ${x^3}dy + (xy - 1)dx = 0,x > 0,y\left( {{1 \over 2}} \right) = 3 - \mathrm{e}$. Then y (1) is equal to

If the solution curve of the differential equation $\frac{d y}{d x}=\frac{x+y-2}{x-y}$ passes through the points $(2,1)$ and $(\mathrm{k}+1,2), \mathrm{k}>0$, then

Let $y=y(x)$ be the solution curve of the differential equation $ \frac{d y}{d x}+\left(\frac{2 x^{2}+11 x+13}{x^{3}+6 x^{2}+11 x+6}\right) y=\frac{(x+3)}{x+1}, x>-1$, which passes through the point $(0,1)$. Then $y(1)$ is equal to :

Let the solution curve $y=y(x)$ of the differential equation $\left(1+\mathrm{e}^{2 x}\right)\left(\frac{\mathrm{d} y}{\mathrm{~d} x}+y\right)=1$ pass through the point $\left(0, \frac{\pi}{2}\right)$. Then, $\lim\limits_{x \rightarrow \infty} \mathrm{e}^{x} y(x)$ is equal to :

Let $y=y(x)$ be the solution curve of the differential equation $ \frac{d y}{d x}+\frac{1}{x^{2}-1} y=\left(\frac{x-1}{x+1}\right)^{1 / 2}$, $x >1$ passing through the point $\left(2, \sqrt{\frac{1}{3}}\right)$. Then $\sqrt{7}\, y(8)$ is equal to :

The differential equation of the family of circles passing through the points $(0,2)$ and $(0,-2)$ is :

Let the solution curve of the differential equation $x \mathrm{~d} y=\left(\sqrt{x^{2}+y^{2}}+y\right) \mathrm{d} x, x>0$, intersect the line $x=1$ at $y=0$ and the line $x=2$ at $y=\alpha$. Then the value of $\alpha$ is :

If $y=y(x), x \in(0, \pi / 2)$ be the solution curve of the differential equation

$\left(\sin ^{2} 2 x\right) \frac{d y}{d x}+\left(8 \sin ^{2} 2 x+2 \sin 4 x\right) y=2 \mathrm{e}^{-4 x}(2 \sin 2 x+\cos 2 x)$,

with $y(\pi / 4)=\mathrm{e}^{-\pi}$, then $y(\pi / 6)$ is equal to :

Let $y=y_{1}(x)$ and $y=y_{2}(x)$ be two distinct solutions of the differential equation $\frac{d y}{d x}=x+y$, with $y_{1}(0)=0$ and $y_{2}(0)=1$ respectively. Then, the number of points of intersection of $y=y_{1}(x)$ and $y=y_{2}(x)$ is

Let the solution curve $y=f(x)$ of the differential equation $ \frac{d y}{d x}+\frac{x y}{x^{2}-1}=\frac{x^{4}+2 x}{\sqrt{1-x^{2}}}$, $x\in(-1,1)$ pass through the origin. Then $\int\limits_{-\frac{\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}} f(x) d x $ is equal to

If ${{dy} \over {dx}} + 2y\tan x = \sin x,\,0 < x < {\pi \over 2}$ and $y\left( {{\pi \over 3}} \right) = 0$, then the maximum value of $y(x)$ is :

Let a smooth curve $y=f(x)$ be such that the slope of the tangent at any point $(x, y)$ on it is directly proportional to $\left(\frac{-y}{x}\right)$. If the curve passes through the points $(1,2)$ and $(8,1)$, then $\left|y\left(\frac{1}{8}\right)\right|$ is equal to

The slope of the tangent to a curve $C: y=y(x)$ at any point $(x, y)$ on it is $\frac{2 \mathrm{e}^{2 x}-6 \mathrm{e}^{-x}+9}{2+9 \mathrm{e}^{-2 x}}$. If $C$ passes through the points $\left(0, \frac{1}{2}+\frac{\pi}{2 \sqrt{2}}\right)$ and $\left(\alpha, \frac{1}{2} \mathrm{e}^{2 \alpha}\right)$, then $\mathrm{e}^{\alpha}$ is equal to :

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

Let ${{dy} \over {dx}} = {{ax - by + a} \over {bx + cy + a}},\,a,b,c \in R$, represents a circle with center ($\alpha$, $\beta$). Then, $\alpha$ + 2$\beta$ is equal to :

If y = y(x) is the solution of the differential equation $\left( {1 + {e^{2x}}} \right){{dy} \over {dx}} + 2\left( {1 + {y^2}} \right){e^x} = 0$ and y (0) = 0, then $6\left( {y'(0) + {{\left( {y\left( {{{\log }_e}\sqrt 3 } \right)} \right)}^2}} \right)$ is equal to

Let the solution curve of the differential equation

$x{{dy} \over {dx}} - y = \sqrt {{y^2} + 16{x^2}} $, $y(1) = 3$ be $y = y(x)$. Then y(2) is equal to:

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

$2y\,{e^{x/{y^2}}}dx + \left( {{y^2} - 4x{e^{x/{y^2}}}} \right)dy = 0$ such that x(1) = 0. Then, x(e) is equal to :

Let the slope of the tangent to a curve y = f(x) at (x, y) be given by 2 $\tan x(\cos x - y)$. If the curve passes through the point $\left( {{\pi \over 4},0} \right)$, then the value of $\int\limits_0^{\pi /2} {y\,dx} $ is equal to :

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

$\left[ {{x \over {\sqrt {{x^2} - {y^2}} }} + {e^{{y \over x}}}} \right]x{{dy} \over {dx}} = x + \left[ {{x \over {\sqrt {{x^2} - {y^2}} }} + {e^{{y \over x}}}} \right]y$

pass through the points (1, 0) and (2$\alpha$, $\alpha$), $\alpha$ > 0. Then $\alpha$ is equal to

Let y = y(x) be the solution of the differential equation $x(1 - {x^2}){{dy} \over {dx}} + (3{x^2}y - y - 4{x^3}) = 0$, $x > 1$, with $y(2) = - 2$. Then y(3) is equal to :

If the solution curve of the differential equation

$(({\tan ^{ - 1}}y) - x)dy = (1 + {y^2})dx$ passes through the point (1, 0), then the abscissa of the point on the curve whose ordinate is tan(1), is

Let ${{dy} \over {dx}} = {{ax - by + a} \over {bx + cy + a}}$, where a, b, c are constants, represent a circle passing through the point (2, 5). Then the shortest distance of the point (11, 6) from this circle is :

If ${{dy} \over {dx}} + {{{2^{x - y}}({2^y} - 1)} \over {{2^x} - 1}} = 0$, x, y > 0, y(1) = 1, then y(2) is equal to :

If $y = y(x)$ is the solution of the differential equation

$x{{dy} \over {dx}} + 2y = x\,{e^x}$, $y(1) = 0$ then the local maximum value

of the function $z(x) = {x^2}y(x) - {e^x},\,x \in R$ is :

If the solution of the differential equation

${{dy} \over {dx}} + {e^x}\left( {{x^2} - 2} \right)y = \left( {{x^2} - 2x} \right)\left( {{x^2} - 2} \right){e^{2x}}$ satisfies $y(0) = 0$, then the value of y(2) is _______________.

If $y = y(x)$ is the solution of the differential equation

$2{x^2}{{dy} \over {dx}} - 2xy + 3{y^2} = 0$ such that $y(e) = {e \over 3}$, then y(1) is equal to :

Let $g:(0,\infty ) \to R$ be a differentiable function such that

$\int {\left( {{{x(\cos x - \sin x)} \over {{e^x} + 1}} + {{g(x)\left( {{e^x} + 1 - x{e^x}} \right)} \over {{{({e^x} + 1)}^2}}}} \right)dx = {{x\,g(x)} \over {{e^x} + 1}} + c} $, for all x > 0, where c is an arbitrary constant. Then :

Let $y = y(x)$ be the solution of the differential equation $(x + 1)y' - y = {e^{3x}}{(x + 1)^2}$, with $y(0) = {1 \over 3}$. Then, the point $x = - {4 \over 3}$ for the curve $y = y(x)$ is :

If the solution curve $y = y(x)$ of the differential equation ${y^2}dx + ({x^2} - xy + {y^2})dy = 0$, which passes through the point (1, 1) and intersects the line $y = \sqrt 3 x$ at the point $(\alpha ,\sqrt 3 \alpha )$, then value of ${\log _e}(\sqrt 3 \alpha )$ is equal to :

If x = x(y) is the solution of the differential equation

$y{{dx} \over {dy}} = 2x + {y^3}(y + 1){e^y},\,x(1) = 0$; then x(e) is equal to :