Let the population of rabbits surviving at time $t$ be governed by the differential equation
$\dfrac{dp(t)}{dt}=\dfrac{1}{2}p(t)-200$. If $p(0)=100$, then $p(t)$ equals:
The area enclosed by the closed curve $\mathcal{C}$ given by the differential equation
$\dfrac{dy}{dx}+\dfrac{x+a}{\,y-2\,}=0,\quad y(1)=0$
is $4\pi$. Let $P$ and $Q$ be the points of intersection of the curve $\mathcal{C}$ with the $y$-axis. If the normals at $P$ and $Q$ on $\mathcal{C}$ intersect the $x$-axis at points $R$ and $S$ respectively, then the length of the line segment $RS$ is:
Let y = y(x) be the solution of the differential equation xdy = (y + x3 cosx)dx with y($\pi$) = 0, then $y\left( {{\pi \over 2}} \right)$ is equal to :
If $y=y(x)$ is the solution curve of the differential equation $\dfrac{dy}{dx}+y\tan x=x\sec x,\ 0\le x\le \dfrac{\pi}{3},\ y(0)=1$, then $y\!\left(\dfrac{\pi}{6}\right)$ is equal to:
If a curve $y=y(x)$ passes through the point $\left(1,\dfrac{\pi}{2}\right)$ and satisfies the differential equation $(7x^{4}\cot y-e^{x}\csc y),\dfrac{dx}{dy}=x^{5},\ x\ge1$, then at $x=2$, the value of $\cos y$ is:
Let the solution curve $y=f(x)$ of the differential equation
$\dfrac{dy}{dx}+\dfrac{xy}{x^{2}-1}=\dfrac{x^{4}+2x}{\sqrt{1-x^{2}}}$, $x\in(-1,1)$,
pass through the origin. Then $\displaystyle \int_{-\sqrt{3}/2}^{\sqrt{3}/2} f(x)\,dx$ is equal to:
Let y = y(x) be solution of the differential equation ${\log _{}}\left( {{{dy} \over {dx}}} \right) = 3x + 4y$, with y(0) = 0.If $y\left( { - {2 \over 3}{{\log }_e}2} \right) = \alpha {\log _e}2$, then the value of $\alpha$ is equal to :
Let $\alpha x = \exp(x^{\beta} y^{\gamma})$ be the solution of the differential equation
$2x^{2}y\,dy - (1 - xy^{2})\,dx = 0,\ x>0,\ y(2)=\sqrt{\log_{e}2}.$
Then $\alpha + \beta - \gamma$ equals:
At present, a firm is manufacturing $2000$ items. It is estimated that the rate of
change of production $P$ w.r.t. additional number of workers $x$ is given by
$\dfrac{dP}{dx}=100-12\sqrt{x}$. If the firm employs $25$ more workers, then the
new level of production of items is :
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 $y = y_{1}(x)$ and $y = y_{2}(x)$ be two distinct solutions of the differential equation $\dfrac{dy}{dx} = x + y$, 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
If $y(x)$ is the solution of the differential equation
$\dfrac{dy}{dx}+\left(\dfrac{2x+1}{x}\right)y=e^{-2x},\ x>0,$ where $y(1)=\dfrac{1}{2}e^{-2}$, then
Let $y=y(x)$ be the solution curve of the differential equation
$x(x^{2}+e^{x})^{2}dy+\big(e^{x}(x-2)y-x^{3}\big)dx=0, x>0,$
passing through the point $(1,0)$. Then $y(2)$ is equal to:
Let y = y(x) be a solution curve of the differential equation $(y + 1){\tan ^2}x\,dx + \tan x\,dy + y\,dx = 0$, $x \in \left( {0,{\pi \over 2}} \right)$. If $\mathop {\lim }\limits_{x \to 0 + } xy(x) = 1$, then the value of $y\left( {{\pi \over 4}} \right)$ is :
Let $x=x(t)$ and $y=y(t)$ be solutions of the differential equations
$\dfrac{dx}{dt}+ax=0$ and $\dfrac{dy}{dt}+by=0$ respectively, $a,b\in\mathbb{R}$.
Given that $x(0)=2$, $y(0)=1$ and $3y(1)=2x(1)$, the value of $t$ for which $x(t)=y(t)$ is:
Let $y = y(x)$ be the solution of the differential equation
$\left(xy - 5x^2\sqrt{1 + x^2}\right)dx + (1 + x^2)dy = 0$, $y(0) = 0$.
Then $y(\sqrt{3})$ is equal to
If the solution curve $f(x,y)=0$ of the differential equation
$(1+\log_e x)\frac{dx}{dy}-x\log_e x=e^y,\; x>0,$
passes through the points $(1,0)$ and $(\alpha,2)$, then $\alpha^\alpha$ is equal to:
Let $y=y(x)$ be the solution of the differential equation
$(x^{2}+1),y'-2xy=(x^{4}+2x^{2}+1)\cos x$, with $y(0)=1$.
Then $\displaystyle \int_{-3}^{3} y(x),dx$ is:
Let the solution curve of the differential equation
$x,dy = \left(\sqrt{x^{2}+y^{2}}+y\right)dx,; x>0,$
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
$$(\sin^{2}2x)\dfrac{dy}{dx}+(8\sin^{2}2x+2\sin 4x)y=2e^{-4x}(2\sin 2x+\cos 2x),$$
with $y(\pi/4)=e^{-\pi}$, then $y(\pi/6)$ is equal to :
$ \text{Let } C(\alpha,\beta) \text{ be the circumcenter of the triangle formed by the lines } 4x+3y=69,; 4y-3x=17,; x+7y=61. $
$ \text{Then } (\alpha-\beta)^2+\alpha+\beta \text{ is equal to:} $
If $y=y(x)$ is the solution curve of the differential equation $(x^2-4)\,dy-(y^2-3y)\,dx=0,\ x>2,\ y(4)=\dfrac{3}{2}$ and the slope of the curve is never zero, then the value of $y(10)$ equals:
$ \text{Let } y=y(x) \text{ be the solution curve of the differential equation } \dfrac{dy}{dx}+\dfrac{1}{x^{2}-1},y=\left(\dfrac{x-1}{x+1}\right)^{1/2},; x>1,\ \text{passing through the point } \left(2,\sqrt{\tfrac{1}{3}}\right). \text{ Then } \sqrt{7},y(8) \text{ is equal to:} $
Let $f:(0,\infty)\to\mathbb{R}$ be a function which is differentiable at all points of its domain and satisfies the condition $x^2 f'(x)=2x f(x)+3$, with $f(1)=4$. Then $2f(2)$ 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 :
Let $y = y(x)$ be the solution of the differential equation
$\sin x \dfrac{dy}{dx} + y \cos x = 4x,\ x \in (0,\pi).$
If $y\left(\dfrac{\pi}{2}\right) = 0$, then $y\left(\dfrac{\pi}{6}\right)$ is equal to :
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 :
Let the solution curve $y = y(x)$ of the differential equation
$\left(1 + e^{2x}\right)\left(\dfrac{dy}{dx} + y\right) = 1$
pass through the point $\left(0, \dfrac{\pi}{2}\right)$.
Then, $\lim_{x \to \infty} e^x y(x)$ is equal to :
$ \text{Let the solution curve } y=y(x) \text{ of the differential equation } (1+e^{2x})!\left(\dfrac{dy}{dx}+y\right)=1 \text{ pass through the point } \left(0,\dfrac{\pi}{2}\right). $
$ \text{Then } \lim_{x\to\infty} e^{x}y(x) \text{ is equal to:} $
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 :
The function $f$ is differentiable and satisfies $x^{2}f(x)-x=4\displaystyle\int_{0}^{x} t f(t),dt$, with $f(1)=\dfrac{2}{3}$. Then $18f(3)$ is equal to:
If a curve passes through the point $(1, -2)$ and has slope of the tangent at any point $(x, y)$ on it as $\dfrac{x^2 - 2y}{x}$, then the curve also passes through the point:
Let $y=y(x)$ be the solution curve of the differential equation
$
\frac{dy}{dx}+\left(\frac{2x^{2}+11x+13}{x^{3}+6x^{2}+11x+6}\right)y=\frac{x+3}{x+1},\quad x>-1,
$
which passes through the point $(0,1)$. Then $y(1)$ is equal to:
Suppose the solution of the differential equation
$ \displaystyle \frac{dy}{dx}=\frac{(2+\alpha)x-\beta y+2}{\beta x-2\alpha y-(\beta\gamma-4\alpha)} $
represents a circle passing through the origin. Then the radius of this circle is:
Let $y = y(x)$ be a solution curve of the differential equation
\[
(1 - x^2 y^2)\,dx = y\,dx + x\,dy.
\]
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 $y = y(x)$ be the solution of the differential equation
$\dfrac{2 + \sin x}{y+1} \cdot \dfrac{dy}{dx} = -\cos x,\; y > 0,\; y(0) = 1.$
If $y(\pi) = a$ and $\dfrac{dy}{dx}$ at $x = \pi$ is $b$,
then the ordered pair $(a,b)$ is equal to :
Let $y=y(x)$ be the solution of the differential equation
$x^{3}\,dy+(xy-1)\,dx=0,\quad x>0,$ with $y\!\left(\dfrac{1}{2}\right)=3-e.$
Then $y(1)$ 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 y = y(x) is the solution of the differential equation, ${{dy} \over {dx}} + 2y\tan x = \sin x,y\left( {{\pi \over 3}} \right) = 0$, then the maximum value of the function y(x) over R is equal to:
f 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 ${{dy} \over {dx}}$ + (tan x) y = sin x, $0 \le x \le {\pi \over 3}$, with y(0) = 0, then $y\left( {{\pi \over 4}} \right)$ 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 $e^{-a}+4a^{2}+a-1$. Then the differential equation whose general solution is $y=c_1f(x)+c_2$, where $c_1$ and $c_2$ are arbitrary constants, is:
Consider the integral $I = \int_0^{10} {{{[x]{e^{[x]}}} \over {{e^{x - 1}}}}dx} $, where [x] denotes the greatest integer less than or equal to x. Then the value of I is equal to :
Let $y=y(x)$ be the solution of the differential equation
$(1+y^{2})e^{\tan x},dx+\cos^{2}x,(1+e^{2\tan x}),dy=0$, $y(0)=1$.
Then $y!\left(\tfrac{\pi}{4}\right)$ is equal to
If a curve y = f(x), passing through the point(1, 2), is the solution of the differential equation, 2x2dy= (2xy + y2)dx, then $f\left( {{1 \over 2}} \right)$ is equal to
Let A = {2, 3, 4, 5, ....., 30} and '$ \simeq $' be an equivalence relation on A $\times$ A, defined by (a, b) $ \simeq $ (c, d), if and only if ad = bc. Then the number of ordered pairs which satisfy this equivalence relation with ordered pair (4, 3) is equal to :
Let $y=y(x)$ be the solution of the differential equation $(x^2+1)^2\dfrac{dy}{dx}+2x(x^2+1)y=1$ such that $y(0)=0$. If $\sqrt{a,y(1)}=\dfrac{\pi}{32}$, then the value of $a$ is:
Let $y=y(x)$ be the solution of the differential equation
\[
\frac{dy}{dx}+\frac{5}{x(x^5+1)}\,y=\frac{(x^5+1)^2}{x^7},\quad x>0.
\]
If $y(1)=2$, then $y(2)$ is equal to:
Let $y=y(x)$ be the solution curve of the differential equation
$\sec y,\dfrac{dy}{dx}+2x\sin y=x^{3}\cos y$, with $y(1)=0$.
Then $y(\sqrt{3})$ 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 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 $y=y(x)$ be the solution curve of the differential equation
$\displaystyle \frac{dy}{dx}=\frac{y}{x}\bigl(1+xy^{2}(1+\log_{e}x)\bigr),\ x>0,\ y(1)=3.$
Then $\displaystyle \frac{y^{2}(x)}{9}$ is equal to:
The population P = P(t) at time 't' of a certain species follows the differential equation ${{dP} \over {dt}}$ = 0.5P – 450. If P(0) = 850, then the time at which population becomes zero is :
If the curve y = y(x) is the solution of the differential equation $2({x^2} + {x^{5/4}})dy - y(x + {x^{1/4}})dx = {2x^{9/4}}dx$, x > 0 which passes through the point $\left( {1,1 - {4 \over 3}{{\log }_e}2} \right)$, then the value of y(16) is equal to :
Let $y=y_1(x)$ and $y=y_2(x)$ be the solution curves of the differential equation $\dfrac{dy}{dx}=y+7$ 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:
If $\sin\left(\dfrac{y}{x}\right)=\log_e|x|+\dfrac{\alpha}{x}$ is a solution of the differential equation
$x\cos\left(\dfrac{y}{x}\right)\dfrac{dy}{dx}=y\cos\left(\dfrac{y}{x}\right)+x$ with $y(1)=\dfrac{\pi}{3}$, then $\alpha^2$ is equal to:
Let y = y(x) be the solution of the differential equation $\cos x(3\sin x + \cos x + 3)dy = (1 + y\sin x(3\sin x + \cos x + 3))dx,0 \le x \le {\pi \over 2},y(0) = 0$. Then, $y\left( {{\pi \over 3}} \right)$ 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 :
Let $y=y(t)$ be a solution of the differential equation $\dfrac{dy}{dt}+\alpha y=\gamma e^{-\beta t}$ where $\alpha>0$, $\beta>0$ and $\gamma>0$.
Then $\displaystyle \lim_{t\to\infty} y(t)$
If $2x = y^{\tfrac{1}{5}} + y^{-\tfrac{1}{5}}$ and
$(x^{2} - 1)\dfrac{d^{2}y}{dx^{2}} + \lambda x\dfrac{dy}{dx} + ky = 0$,
then $\lambda + k$ is equal to:
Three rotten apples are accidentally mixed with fifteen good apples. Assuming the random
variable $x$ to be the number of rotten apples in a draw of two apples, the variance of $x$ is:
The solution curve of the differential equation $2y\dfrac{dy}{dx}+3=5\dfrac{dy}{dx}$, passing through the point $(0,1)$, is a conic whose vertex lies on the line:
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 $\displaystyle \int_{0}^{x}\sqrt{1-\big(y'(t)\big)^{2}},dt=\int_{0}^{x}y(t),dt,\ 0\le x\le 3,\ y\ge0,\ y(0)=0$.
Then at $x=2$, $,y''+y+1$ is equal to:
If $\cos x{{dy} \over {dx}} - y\sin x = 6x$, (0 < x < ${\pi \over 2}$)
and $y\left( {{\pi \over 3}} \right)$ = 0 then $y\left( {{\pi \over 6}} \right)$ is equal to :
If a curve $y = f(x)$ passes through the point $(1,-1)$ and satisfies the differential equation
$ y(1+xy),dx = x,dy $,
then $ f\left(-\dfrac{1}{2}\right) $ is equal to:
Let $x=x(y)$ be the solution of the differential equation
$2(y+2)\log_e(y+2)\,dx+\big(x+4-2\log_e(y+2)\big)\,dy=0,\quad y>-1$
with $x\big(e^{4}-2\big)=1$. Then $x\big(e^{9}-2\big)$ is equal to:
Let $f:[1, \infty) \rightarrow[2, \infty)$ be a differentiable function. If $10 \int_1^1 f(\mathrm{t}) \mathrm{dt}=5 x f(x)-x^5-9$ for all $x \geqslant 1$, then the value of $f(3)$ is :
Let $y=y(x)$ be the solution of the differential equation
$$x\log_e x \,\frac{dy}{dx}+y=x^2\log_e x,\quad (x>1).$$
If $y(2)=2$, then $y(e)$ is equal to:
If ${{dy} \over {dx}} = {{{2^x}y + {2^y}{{.2}^x}} \over {{2^x} + {2^{x + y}}{{\log }_e}2}}$, y(0) = 0, then for y = 1, the value of x lies in the interval :
The temperature $T(t)$ of a body at time $t=0$ is $160^\circ\!F$ and it decreases
continuously as per the differential equation $\dfrac{dT}{dt}=-K(T-80)$,
where $K$ is a positive constant. If $T(15)=120^\circ\!F$, then $T(45)$ is:
Let y = y(x) be the solution of the differential equation $x\tan \left( {{y \over x}} \right)dy = \left( {y\tan \left( {{y \over x}} \right) - x} \right)dx$, $ - 1 \le x \le 1$, $y\left( {{1 \over 2}} \right) = {\pi \over 6}$. Then the area of the region bounded by the curves x = 0, $x = {1 \over {\sqrt 2 }}$ and y = y(x) in the upper half plane is :
If $y=y(x)$ is the solution of the differential equation
$x\dfrac{dy}{dx}+2y=x^{2}$, satisfying $y(1)=1$, then $y\!\left(\dfrac{1}{2}\right)$ is equal to:
Let y = y(x) be the solution of the differential equation ${e^x}\sqrt {1 - {y^2}} dx + \left( {{y \over x}} \right)dy = 0$, y(1) = $-$1. Then the value of (y(3))2 is equal to :
Let the solution curve $y=y(x)$ of the differential equation
\[
\frac{dy}{dx}=\frac{3x^5\tan^{-1}(x^3)}{(1+x^6)^{3/2}}\, y = 2x \exp\left\{\frac{x^3-\tan^{-1}(x^3)}{\sqrt{1+x^6}}\right\}
\]
pass through the origin. Then $y(1)$ is equal to:
If $y=y(x)$ is the solution of the differential equation $\dfrac{dy}{dx}=(\tan x-y)\sec^{2}x,\ x\in\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)$, such that $y(0)=0$, then $y!\left(-\dfrac{\pi}{4}\right)$ 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
If y = y(x) is the solution curve of the differential equation ${x^2}dy + \left( {y - {1 \over x}} \right)dx = 0$ ; x > 0 and y(1) = 1, then $y\left( {{1 \over 2}} \right)$ is equal to :
Let y = y(x) be the solution of the differential equation, xy'- y = x2(xcosx + sinx), x > 0. if y ($\pi $) = $\pi $ then $y''\left( {{\pi \over 2}} \right) + y\left( {{\pi \over 2}} \right)$ is equal to :
Let $x=x(y)$ be the solution of the differential equation $y^2\,dx+\left(x-\dfrac{1}{y}\right)dy=0$. If $x(1)=1$, then $x\!\left(\dfrac{1}{2}\right)$ is:
Let $g$ be a differentiable function such that
$\displaystyle \int_0^x g(t),dt = x - \int_0^x t g(t),dt,; x \ge 0$
and let $y = y(x)$ satisfy the differential equation
$\dfrac{dy}{dx} - y \tan x = 2(x + 1)\sec x, g(x),; x \in \left[0, \dfrac{\pi}{2}\right).$
If $y(0) = 0$, then $y\left(\dfrac{\pi}{3}\right)$ is equal to
Let y = y(x) be the solution of the differential equation $\cos e{c^2}xdy + 2dx = (1 + y\cos 2x)\cos e{c^2}xdx$, with $y\left( {{\pi \over 4}} \right) = 0$. Then, the value of ${(y(0) + 1)^2}$ is equal to :
For $x \in \mathbb{R},\ x \ne 0$, if $y(x)$ is a differentiable function such that
$x \int_{1}^{x} y(t)\,dt = (x+1) \int_{1}^{x} t\,y(t)\,dt,$
then $y(x)$ equals: (where $C$ is a constant.)
Let $y=y(x)$ be the solution of the differential equation
$\dfrac{dy}{dx}+y\tan x=2x+x^{2}\tan x,\ x\in\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)$, such that $y(0)=1$. Then:
If $x=f(y)$ is the solution of the differential equation $(1+y^{2})+\big(x-2e^{\tan^{-1}y}\big)\dfrac{dy}{dx}=0,\ y\in\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)$ with $f(0)=1$, then $f\!\left(\dfrac{1}{\sqrt{3}}\right)$ is:
Let $\alpha$ be a non-zero real number. Suppose $f:\mathbf{R}\to\mathbf{R}$ is a differentiable function such that
$f(0)=2$ and $\displaystyle \lim_{x\to -\infty} f(x)=1$.
If $f'(x)=\alpha f(x)+3$, for all $x\in\mathbf{R}$, then $f(-\log_{e}2)$ is equal to:
If y = y(x) is the solution of the differential equation ${{5 + {e^x}} \over {2 + y}}.{{dy} \over {dx}} + {e^x} = 0$ satisfyingy(0) = 1, then a value of y(loge13) is :
Let $f:[0,1]\to\mathbb{R}$ be such that $f(xy)=f(x)\,f(y)$ for all $x,y\in[0,1]$, and $f(0)\ne 0$.
If $y=v(x)$ satisfies the differential equation $\dfrac{dy}{dx}=f(x)$ with $y(0)=1$, then $y\!\left(\dfrac{1}{4}\right)+y\!\left(\dfrac{3}{4}\right)$ is equal to:
Let y = y(x) be the solution of the differential equation ${{dy} \over {dx}} = 1 + x{e^{y - x}}, - \sqrt 2 < x < \sqrt 2 ,y(0) = 0$ then, the minimum value of $y(x),x \in \left( { - \sqrt 2 ,\sqrt 2 } \right)$ is equal to :
Let y = y(x) be the solution of the differentialequationcosx${{dy} \over {dx}}$ + 2ysinx = sin2x, x $ \in $ $\left( {0,{\pi \over 2}} \right)$.Ify$\left( {{\pi \over 3}} \right)$ = 0, then y$\left( {{\pi \over 4}} \right)$ is equal to :
Let a curve $y=f(x)$ pass through the points $(0,5)$ and $(\log_e 2,\,k)$. If the curve satisfies the differential equation $2(3+y)e^{2x}\,dx-(7+e^{2x})\,dy=0$, then $k$ is equal to:
Let $y=y(x)$ be the solution of the differential equation $(3y^{2}-5x^{2})\,y\,dx+2x\,(x^{2}-y^{2})\,dy=0$ such that $y(1)=1$. Then $\left|(y(2))^{3}-12y(2)\right|$ is equal to:
Let $f:[0,\infty)\to\mathbb{R}$ be a differentiable function such that
$f(x)=1-2x+\displaystyle\int_{0}^{x}e^{,x-t}f(t),dt$ for all $x\in[0,\infty)$.
Then the area of the region bounded by $y=f(x)$ and the coordinate axes is
If the solution $y = y(x)$ of the differential equation
$(x^{4}+2x^{3}+3x^{2}+2x+2)\,dy-(2x^{2}+2x+3)\,dx=0$
satisfies $y(-1)=-\dfrac{\pi}{4}$, then $y(0)$ is equal to:
If \(\dfrac{dy}{dx}+\dfrac{3}{\cos^2 x}\,y=\dfrac{1}{\cos^2 x},\ x\in\left(-\dfrac{\pi}{3},\dfrac{\pi}{3}\right)\) and \(y\!\left(\dfrac{\pi}{4}\right)=\dfrac{4}{3}\), then \(y\!\left(-\dfrac{\pi}{4}\right)\) equals:
Let $f(x)=
\begin{cases}
x^{3}-x^{2}+10x-7, & x\le 1,\\
-2x+\log_{2}(b^{2}-4), & x>1.
\end{cases}$
Then the set of all values of $b$ for which $f(x)$ has maximum value at $x=1$ is:
The general solution of the differential equation$\sqrt {1 + {x^2} + {y^2} + {x^2}{y^2}} $ + xy${{dy} \over {dx}}$ = 0 is : (where C is a constant of integration)
Let $f$ be a differentiable function such that
$f'(x) = 7 - \dfrac{3}{4}\,\dfrac{f(x)}{x}$, for $x>0$, and $f(1)\neq 4$.
Then $\displaystyle \lim_{x\to 0} x\,f\!\left(\dfrac{1}{x}\right)$ equals: