Variable Separable Method

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JEE Main 2016

If a curve $y = f (x)$ passes through the point $(1, - 1)$ and satisfies the differential equation, $y (1 + x y) d x = x d y$ , then $f (- \frac{1}{2})$ is equal to :

\frac{2}{5}

(B)

\frac{4}{5}

(C)

- \frac{2}{5}

(D)

- \frac{4}{5}

JEE Main 2014

LEVELJEE Main

Let the population of rabbits surviving at time $t$ be governed by the differential equation $\frac{d p ( t )}{d t} = \frac{1}{2} p (t) - 200$ . If $p (0) = 100$ , then $p (t)$ equals:

(A)

600 - 500 e^{t /2}

(B)

400 - 300 e^{- t /2}

(C)

400 - 300 e^{t /2}

(D)

300 - 200 e^{- t /2}

JEE Main 2013

LEVELJEE Main

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 $\frac{d P}{d x} = 100 - 12 x$ . If the firm employs 25 more workers, then the new level of production of items is

The population $p (t)$ at time $t$ of a certain mouse species satisfies the differential equation $\frac{d p ( t )}{d t} = 0.5 p (t) - 450$ . If $p (0) = 850$ , then the time at which the population becomes zero is :

(A)

2 ln 18

(B)

ln 9

(C)

\frac{1}{2} ln 18

(D)

ln 18

JEE Main 2011

LEVELJEE Main

If $\frac{d y}{d x} = y + 3 > 0$ and $y (0) = 2$ , then $y (ln 2)$ is equal to :

Let $I$ be the purchase value of an equipment and $V (t)$ be the value after it has been used for $t$ years. The value $V (t)$ depreciates at a rate given by differential equation $\frac{d V ( t )}{d t} = - k (T - t)$ , where $k > 0$ is a constant and $T$ is the total life in years of the equipment. Then the scrap value $V (T)$ of the equipment is

(A)

I - \frac{k T ^{2}}{2}

(B)

I - \frac{k ( T - t ) ^{2}}{2}

(C)

e^{- k T}

(D)

T^{2} - \frac{1}{k}

JEE Main 2007

LEVELJEE Main

The normal to a curve at $P (x, y)$ meets the x-axis at $G$ . If the distance of $G$ from the origin is twice the abscissa of $P$ , then the curve is a

A function $y = f (x)$ has a second order derivative $f^{''} (x) = 6 (x - 1)$ . If its graph passes through the point $(2, 1)$ and at that point the tangent to the graph is $y = 3 x - 5$ , then the function is

Solution of the differential equation $y d x + (x + x^{2} y) d y = 0$ is

(A)

lo g y = C x

(B)

- \frac{1}{x y} + lo g y = C

(C)

\frac{1}{x y} + lo g y = C

(D)

- \frac{1}{x y} = C

JEE Main 2002

LEVELBoard

$f (x)$ and $g (x)$ are two differentiable functions on $[0, 2]$ such that $f^{''} (x) - g^{''} (x) = 0, f^{'} (1) = 2 g^{'} (1) = 4, f (2) = 3 g (2) = 9$ then $f (x) - g (x)$ at $x = 3/2$ is

The solution of the equation $\frac{d ^{2} y}{d x ^{2}} = e^{- 2 x}$

(A)

\frac{e ^{- 2 x}}{4}

(B)

\frac{e ^{- 2 x}}{4} + c x + d

(C)

\frac{1}{4} e^{- 2 x} + c x^{2} + d

(D)

\frac{1}{4} e^{- 4 x} + c x + d

JEE Advanced 2014

LEVELJEE Main

Let $f : [0, 2] \to R$ be a function which is continuous on $[0, 2]$ and is differentiable on $(0, 2)$ with $f (0) = 1$ . Let $F (x) = \int_{0}^{x^{2}} f (t) d t$ for $x \in [0, 2]$ . If $F^{'} (x) = f^{'} (x)$ for all $x \in (0, 2)$ , then $f (2)$ equals

(A)

e^{2} - 1

(B)

e^{4} - 1

(C)

e - 1

(D)

e^{4}

JEE Advanced 2008

LEVELJEE Advanced

Let a solution $y = y (x)$ of the differential equation $x x^{2} - 1 d y - y y^{2} - 1 d x = 0$ satisfy $y (2) = \frac{2}{3}$ . STATEMENT-1 : $y (x) = sec (sec^{- 1} x - \frac{π}{6})$ and STATEMENT-2 : $y (x)$ is given by $\frac{1}{y} = \frac{2 3}{x} - 1 - \frac{1}{x ^{2}}$

(A)

STATEMENT - 1 is True, STATEMENT - 2 is True; STATEMENT - 2 is a correct explanation for STATEMENT - 1

(B)

STATEMENT - 1 is True, STATEMENT - 2 is True; STATEMENT - 2 is NOT a correct explanation for STATEMENT - 1

(C)

STATEMENT - 1 is True, STATEMENT - 2 is False

(D)

STATEMENT - 1 is False, STATEMENT - 2 is True

JEE Advanced 2006

LEVELJEE Advanced

A curve $y = f (x)$ passes through (1, 1) and at $P (x, y)$ , tangent cuts the x-axis and y-axis at A and B respectively such that $B P : A P = 3 : 1$ , then

* Multiple Correct Options

(A)

equation of curve is

x y^{'} - 3 y = 0

(B)

normal at (1, 1) is

x + 3 y = 4

(C)

curve passes through (2, 1/8)

(D)

equation of curve is

x y^{'} + 3 y = 0

JEE Advanced 2005

LEVELJEE Main

The differential equation $\frac{d y}{d x} = \frac{1 - y ^{2}}{y}$ determines a family of circles with

(A)

variable radii and a fixed centre at (0, 1)

(B)

variable radii and a fixed centre at (0, -1)

(C)

fixed radius 1 and variable centres along the x-axis

(D)

fixed radius 1 and variable centres along the y-axis

JEE Advanced 2004

LEVELJEE Main

If $y = y (x)$ and $\frac{2 + s i n x}{y + 1} (\frac{d y}{d x}) = - cos x, y (0) = 1$ , then $y (\frac{π}{2})$ equals

If length of tangent at any point on the curve $y = f (x)$ intercepted between the point and the x-axis is of length 1. Find the equation of the curve.

JEE Advanced 2001

LEVELJEE Advanced

A hemispherical tank of radius 2 metres is initially full of water and has an outlet of $12 cm^{2}$ cross-sectional area at the bottom. The outlet is opened at some instant. The flow through the outlet is according to the law $v (t) = 0.6 2 g h (t)$ , where $v (t)$ and $h (t)$ are respectively the velocity of the flow through the outlet and the height of water level above the outlet at time t, and g is the acceleration due to gravity. Find the time it takes to empty the tank.

JEE Advanced 1998

LEVELJEE Advanced

A curve $C$ has the property that if the tangent drawn at any point $P$ on $C$ meets the co-ordinate axes at $A$ and $B$ , then $P$ is the mid-point of $A B$ . The curve passes through the point $(1, 1)$ . Determine the equation of the curve.

JEE Advanced 1996

LEVELJEE Advanced

A curve $y = f (x)$ passes through the point $P (1, 1)$ . The normal to the curve at $P$ is $a (y - 1) + (x - 1) = 0$ . If the slope of the tangent at any point on the curve is proportional to the ordinate of the point, determine the equation of the curve. Also obtain the area bounded by the $y$ -axis, the curve and the normal to the curve at $P$ .

JEE Advanced 1996

LEVELJEE Main

Determine the equation of the curve passing through the origin, in the form $y = f (x)$ , which satisfies the differential equation $\frac{d y}{d x} = sin (10 x + 6 y)$ .

JEE Advanced 1994

LEVELJEE Advanced

A normal is drawn at a point $P (x, y)$ of a curve. It meets the x-axis at Q. If PQ is of constant length $k$ , then show that the differential equation describing such curves is $y \frac{d y}{d x} = \pm k^{2} - y^{2}$ . Find the equation of such a curve passing through (0, k).

JEE Advanced 2006

LEVELJEE Advanced

Match the following :

List-I

(P)

\int_{0}^{π /2} (sin x)^{c o s x} (cos x cot x - lo g (sin x)^{s i n x}) d x

(Q)

Area bounded by

- 4 y^{2} = x

and

x - 1 = - 5 y^{2}

(R)

Cosine of the angle of intersection of curves

y = 3^{x - 1} lo g x

and

y = x^{x} - 1

(S)

Let

\frac{d y}{d x} = \frac{6}{x + y}

where

y (0) = 0

then value of

y

when

x + y = 6

List-II

(1)

(2)

(3)

6 ln 2

(4)

4/3

JEE Main 2017

LEVELBoard

If $(2 + sin x) \frac{d y}{d x} + (y + 1) cos x = 0$ and $y (0) = 1$ , then $y (\frac{π}{2})$ is equal to:

(A)

\frac{1}{3}

(B)

- \frac{2}{3}

(C)

- \frac{1}{3}

(D)

\frac{4}{3}

JEE Main 2019 (9 January)

LEVELJEE Main

Let $f : [0, 1] \to R$ be such that $f (x y) = f (x) . f (y)$ for all $x, y \in [0, 1]$ , and $f (0) \neq = 0$ . If $y = y (x)$ satisfies the differential equation, $\frac{d y}{d x} = f (x)$ with $y (0) = 1$ , then $y (1/4) + y (3/4)$ is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If a curve y=f(x) passes through the point (1,−1) and satisfies the differential equation, y(1+xy)dx=xdy, then f(−21​) is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let the population of rabbits surviving at time t be governed by the differential equation dtdp(t)​=21​p(t)−200. If p(0)=100, then p(t) equals:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If dxdy​=y+3>0 and y(0)=2, then y(ln2) is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The normal to a curve at P(x,y) meets the x-axis at G. If the distance of G from the origin is twice the abscissa of P, then the curve is a

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Solution of the differential equation ydx+(x+x2y)dy=0 is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x) and g(x) are two differentiable functions on [0,2] such that f′′(x)−g′′(x)=0,f′(1)=2g′(1)=4,f(2)=3g(2)=9 then f(x)−g(x) at x=3/2 is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The solution of the equation dx2d2y​=e−2x

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A curve y=f(x) passes through (1, 1) and at P(x,y), tangent cuts the x-axis and y-axis at A and B respectively such that BP:AP=3:1, then

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The differential equation dxdy​=y1−y2​​ determines a family of circles with

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If y=y(x) and y+12+sinx​(dxdy​)=−cosx,y(0)=1, then y(2π​) equals

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If length of tangent at any point on the curve y=f(x) intercepted between the point and the x-axis is of length 1. Find the equation of the curve.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Determine the equation of the curve passing through the origin, in the form y=f(x), which satisfies the differential equation dxdy​=sin(10x+6y).

Match the following :

List-I

List-II

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If (2+sinx)dxdy​+(y+1)cosx=0 and y(0)=1, then y(2π​) is equal to:

If a curve $y = f (x)$ passes through the point $(1, - 1)$ and satisfies the differential equation, $y (1 + x y) d x = x d y$ , then $f (- \frac{1}{2})$ is equal to :

Let the population of rabbits surviving at time $t$ be governed by the differential equation $\frac{d p ( t )}{d t} = \frac{1}{2} p (t) - 200$ . If $p (0) = 100$ , then $p (t)$ equals:

If $\frac{d y}{d x} = y + 3 > 0$ and $y (0) = 2$ , then $y (ln 2)$ is equal to :

The normal to a curve at $P (x, y)$ meets the x-axis at $G$ . If the distance of $G$ from the origin is twice the abscissa of $P$ , then the curve is a

Solution of the differential equation $y d x + (x + x^{2} y) d y = 0$ is

$f (x)$ and $g (x)$ are two differentiable functions on $[0, 2]$ such that $f^{''} (x) - g^{''} (x) = 0, f^{'} (1) = 2 g^{'} (1) = 4, f (2) = 3 g (2) = 9$ then $f (x) - g (x)$ at $x = 3/2$ is

The solution of the equation $\frac{d ^{2} y}{d x ^{2}} = e^{- 2 x}$

A curve $y = f (x)$ passes through (1, 1) and at $P (x, y)$ , tangent cuts the x-axis and y-axis at A and B respectively such that $B P : A P = 3 : 1$ , then

The differential equation $\frac{d y}{d x} = \frac{1 - y ^{2}}{y}$ determines a family of circles with

If $y = y (x)$ and $\frac{2 + s i n x}{y + 1} (\frac{d y}{d x}) = - cos x, y (0) = 1$ , then $y (\frac{π}{2})$ equals

If length of tangent at any point on the curve $y = f (x)$ intercepted between the point and the x-axis is of length 1. Find the equation of the curve.

Determine the equation of the curve passing through the origin, in the form $y = f (x)$ , which satisfies the differential equation $\frac{d y}{d x} = sin (10 x + 6 y)$ .

If $(2 + sin x) \frac{d y}{d x} + (y + 1) cos x = 0$ and $y (0) = 1$ , then $y (\frac{π}{2})$ is equal to: