Linear Differential Equations

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JEE Main 2015

LEVELJEE Main

Let $y (x)$ be the solution of the differential equation $(x lo g x) \frac{d y}{d x} + y = 2 x lo g x, (x \geq 1)$ . Then $y (e)$ is equal to:

Solution of the differential equation $cos x d y = y (sin x - y) d x, 0 < x < \frac{π}{2}$ is

(A)

y sec x = tan x + c

(B)

y tan x = sec x + c

(C)

tan x = (sec x + c) y

(D)

sec x = (tan x + c) y

JEE Main 2003

LEVELJEE Main

Let $f (x)$ be a function satisfying $f^{''} (x) = f (x)$ with $f (0) = 1$ and $g (x)$ be a function that satisfies $f (x) + g (x) = x^{2}$ . Then the value of the integral $\int_{0}^{1} f (x) g (x) d x$ , is

(A)

e + \frac{e ^{2}}{2} + \frac{5}{2}

(B)

e - \frac{e ^{2}}{2} - \frac{5}{2}

(C)

e + \frac{e ^{2}}{2} - \frac{3}{2}

(D)

e - \frac{e ^{2}}{2} - \frac{3}{2}

JEE Main 2003

LEVELJEE Main

The solution of the differential equation $(1 + y^{2}) + (x - e^{t a n^{- 1} y}) \frac{d y}{d x} = 0$ , is

(A)

x e^{2 t a n^{- 1} y} = e^{t a n^{- 1} y} + k

(B)

(x - 2) = k e^{2 t a n^{- 1} y}

(C)

2 x e^{t a n^{- 1} y} = e^{2 t a n^{- 1} y} + k

(D)

x e^{t a n^{- 1} y} = tan^{- 1} y + k

JEE Advanced 2016

LEVELJEE Main

Let $f : (0, \infty) \to R$ be a differentiable function such that $f^{'} (x) = 2 - \frac{f ( x )}{x}$ for all $x \in (0, \infty)$ and $f (1) \neq = 1$ . Then

(A)

lim_{x \to 0^{+}} f^{'} (\frac{1}{x}) = 1

(B)

lim_{x \to 0^{+}} x f (\frac{1}{x}) = 2

(C)

lim_{x \to 0^{+}} x^{2} f^{''} (x) = 0

(D)

|f(x)| \leq 2 \text{ for all } x \in (0, 2)

JEE Advanced 2016

LEVELJEE Advanced

A solution curve of the differential equation $(x^{2} + x y + 4 x + 2 y + 4) \frac{d y}{d x} - y^{2} = 0, x > 0$ , passes through the point (1, 3). Then the solution curve

* Multiple Correct Options

(A)

intersects

y = x + 2

exactly at one point

(B)

intersects

y = x + 2

exactly at two points

(C)

intersects

y = (x + 2)^{2}

(D)

does NOT intersect

y = (x + 3)^{2}

JEE Advanced 2015

LEVELJEE Main

Let $y (x)$ be a solution of the differential equation $(1 + e^{x}) y^{'} + y e^{x} = 1$ . If $y (0) = 2$ , then which of the following statement is (are) true?

* Multiple Correct Options

y(x) has a critical point in the interval (-1, 0)

(D)

y(x) has no critical point in the interval (-1, 0)

JEE Advanced 2014

LEVELJEE Main

The function $y = f (x)$ is the solution of the differential equation $\frac{d y}{d x} + \frac{x y}{x ^{2} - 1} = \frac{x ^{4} + 2 x}{1 - x ^{2}}$ in $(- 1, 1)$ satisfying $f (0) = 0$ . Then $\int_{- 3 /2}^{3 /2} f (x) d x$ is

(A)

\frac{π}{3} - \frac{3}{2}

(B)

\frac{π}{3} - \frac{3}{4}

(C)

\frac{π}{6} - \frac{3}{4}

(D)

\frac{π}{6} - \frac{3}{2}

JEE Advanced 2012

LEVELJEE Main

If $y (x)$ satisfies the differential equation $y^{'} - y tan x = 2 x sec x$ and $y (0) = 0$ , then

* Multiple Correct Options

(A)

y(\frac{\pi}{4}) = \frac{\pi^2}{8\sqrt{2}}

(B)

y'(\frac{\pi}{4}) = \frac{\pi^2}{18}

(C)

y(\frac{\pi}{3}) = \frac{\pi^2}{9}

(D)

y'(\frac{\pi}{3}) = \frac{4\pi}{3} + \frac{2\pi^2}{3\sqrt{3}}

JEE Advanced 2007

LEVELJEE Main

Let $f (x)$ be differentiable on the interval $(0, \infty)$ such that $f (1) = 1$ , and $lim_{t \to x} \frac{t ^{2} f ( x ) - x ^{2} f ( t )}{t - x} = 1$ for each $x > 0$ . Then $f (x)$ is

(A)

\frac{1}{3 x} + \frac{2 x ^{2}}{3}

(B)

- \frac{1}{3 x} + \frac{4 x ^{2}}{3}

(C)

- \frac{1}{x} + \frac{2}{x ^{2}}

(D)

\frac{1}{x}

JEE Advanced 2005

LEVELJEE Main

For the primitive integral equation $y d x + y^{2} d y = x d y$ ; $x \in R, y > 0, y (1) = 1$ , then $y (- 3)$ is

If $y (t)$ is a solution of $(1 + t) \frac{d y}{d t} - t y = 1$ and $y (0) = - 1$ , then $y (1)$ is equal to

A solution of the differential equation $(\frac{d y}{d x})^{2} - x \frac{d y}{d x} + y = 0$ is

Let $f : [1, \infty) \to [2, \infty)$ be a differentiable function such that $f (1) = 2$ . If $6 \int_{1}^{x} f (t) d t = 3 x f (x) - x^{3}$ for all $x \geq 1$ , then the value of $f (2)$ is .........

JEE Advanced 2011

LEVELJEE Main

Let $y^{'} (x) + y (x) g^{'} (x) = g (x) g^{'} (x)$ , $y (0) = 0$ , $x \in R$ , where $f^{'} (x)$ denotes $\frac{df ( x )}{d x}$ and $g (x)$ is a given non-constant differentiable function on $R$ with $g (0) = g (2) = 0$ . Then the value of $y (2)$ is

JEE Advanced 2010

LEVELJEE Main

Let $f$ be a real-valued differentiable function on $R$ such that $f (1) = 1$ . If the $y$ -intercept of the tangent at any point $P (x, y)$ on the curve $y = f (x)$ is equal to the cube of the abscissa of $P$ , find the value of $f (- 3)$ .

JEE Advanced 2009

LEVELJEE Main

Let $f : R \to R$ be a continuous function which satisfies $f (x) = \int_{0}^{x} f (t) d t$ . Then the value of $f (ln 5)$ is

JEE Advanced 2004

LEVELJEE Advanced

A curve 'C' passes through (2, 0) and the slope at (x, y) as $\frac{( x + 1 ) ^{2} + ( y - 3 )}{x + 1}$ . Find the equation of the curve. Find the area bounded by curve and x-axis in fourth quadrant.

JEE Advanced 2001

LEVELJEE Main

Let $f (x), x \geq 0$ , be a non-negative continuous function, and let $F (x) = \int_{0}^{x} f (t) d t, x \geq 0$ . If for some $c > 0, f (x) \leq c F (x)$ for all $x \geq 0$ , then show that $f (x) = 0$ for all $x \geq 0$ .

JEE Advanced 2000

LEVELJEE Main

A country has a food deficit of 10%. Its population grows continuously at a rate of 3% per year. Its annual food production every year is 4% more than that of the last year. Assuming that the average food requirement per person remains constant, prove that the country will become self-sufficient in food after n years, where n is the smallest integer bigger than or equal to $\frac{l n 10 - l n 9}{l n ( 1.04 ) - 0.03}$ .

JEE Advanced 1997

LEVELJEE Main

Let $u (x)$ and $v (x)$ satisfy the differential equation $\frac{d u}{d x} + p (x) u = f (x)$ and $\frac{d v}{d x} + p (x) v = g (x)$ , where $p (x), f (x)$ and $g (x)$ are continuous functions. If $u (x_{1}) > v (x_{1})$ for some $x_{1}$ and $f (x) > g (x)$ for all $x > x_{1}$ , prove that any point $(x, y)$ where $x > x_{1}$ , does not satisfy the equations $y = u (x)$ and $y = v (x)$ .

JEE Advanced 2009

LEVELJEE Advanced

Match the statements/expressions given in Column-I with the values given in Column-II.

List-I

(P)

The number of solutions of the equation

x e^{s i n x} - cos x = 0

in the interval

(0, \frac{π}{2})

(Q)

Value(s) of

k

for which the planes

k x + 4 y + z = 0

4 x + k y + 2 z = 0

and

2 x + 2 y + z = 0

intersect in a straight line

(R)

Value(s) of

k

for which

∣ x - 1∣ + ∣ x - 2∣ + ∣ x + 1∣ + ∣ x + 2∣ = 4 k

has integer solution(s)

(S)

y^{'} = y + 1

and

y (0) = 1

, then value(s) of

y (ln 2)

List-II

JEE Main 2018 (Paper 1)

LEVELJEE Main

Let $y = y (x)$ be the solution of the differential equation $sin x \frac{d y}{d x} + y cos x = 4 x, x \in (0, π)$ . If $y (π /2) = 0$ , then $y (π /6)$ is equal to :

(A)

-\frac{4}{9}\pi^2

(B)

\frac{4}{9 3} π^{2}

(C)

-\frac{8}{9\sqrt{3}}\pi^2

(D)

-\frac{8}{9}\pi^2

JEE Main 2018 (15 April Shift 1)

LEVELJEE Main

Let $y = y (x)$ be the solution of the differential equation $\frac{d y}{d x} + 2 y = f (x)$ , where $f (x) = {1, 0, x \in [0, 1] otherwise$ . If $y (0) = 0$ , then $y (3/2)$ is :

(A)

\frac{e ^{2} - 1}{2 e ^{3}}

(B)

\frac{e ^{2} - 1}{e ^{3}}

(C)

\frac{e ^{2} + 1}{2 e ^{4}}

(D)

\frac{1}{2 e}

JEE Main 2019 (9 January)

LEVELJEE Main

If $y = y (x)$ is the solution of the differential equation $x \frac{d y}{d x} + 2 y = x^{2}$ satisfying $y (1) = 1$ , then $y (1/2)$ is equal to :

(A)

\frac{7}{64}

(B)

\frac{13}{16}

(C)

\frac{49}{16}

(D)

\frac{1}{4}

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let y(x) be the solution of the differential equation (xlogx)dxdy​+y=2xlogx,(x≥1). Then y(e) is equal to:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Solution of the differential equation cosxdy=y(sinx−y)dx,0<x<2π​ is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f(x) be a function satisfying f′′(x)=f(x) with f(0)=1 and g(x) be a function that satisfies f(x)+g(x)=x2. Then the value of the integral ∫01​f(x)g(x)dx, is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The solution of the differential equation (1+y2)+(x−etan−1y)dxdy​=0, is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f:(0,∞)→R be a differentiable function such that f′(x)=2−xf(x)​ for all x∈(0,∞) and f(1)=1. Then

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A solution curve of the differential equation (x2+xy+4x+2y+4)dxdy​−y2=0,x>0, passes through the point (1, 3). Then the solution curve

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let y(x) be a solution of the differential equation (1+ex)y′+yex=1. If y(0)=2, then which of the following statement is (are) true?

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The function y=f(x) is the solution of the differential equation dxdy​+x2−1xy​=1−x2​x4+2x​ in (−1,1) satisfying f(0)=0. Then ∫−3​/23​/2​f(x)dx is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If y(x) satisfies the differential equation y′−ytanx=2xsecx and y(0)=0, then

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f(x) be differentiable on the interval (0,∞) such that f(1)=1, and limt→x​t−xt2f(x)−x2f(t)​=1 for each x>0. Then f(x) is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } For the primitive integral equation ydx+y2dy=xdy; x∈R,y>0,y(1)=1, then y(−3) is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If y(t) is a solution of (1+t)dtdy​−ty=1 and y(0)=−1, then y(1) is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A solution of the differential equation (dxdy​)2−xdxdy​+y=0 is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f:[1,∞)→[2,∞) be a differentiable function such that f(1)=2. If 6∫1x​f(t)dt=3xf(x)−x3 for all x≥1, then the value of f(2) is .........

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f:R→R be a continuous function which satisfies f(x)=∫0x​f(t)dt. Then the value of f(ln5) is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f(x),x≥0, be a non-negative continuous function, and let F(x)=∫0x​f(t)dt,x≥0. If for some c>0,f(x)≤cF(x) for all x≥0, then show that f(x)=0 for all x≥0.

Match the statements/expressions given in Column-I with the values given in Column-II.

List-I

List-II

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let y=y(x) be the solution of the differential equation sinxdxdy​+ycosx=4x,x∈(0,π). If y(π/2)=0, then y(π/6) is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let y=y(x) be the solution of the differential equation dxdy​+2y=f(x), where f(x)={1,0,​x∈[0,1]otherwise​. If y(0)=0, then y(3/2) is :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If y=y(x) is the solution of the differential equation xdxdy​+2y=x2 satisfying y(1)=1, then y(1/2) is equal to :

Let $y (x)$ be the solution of the differential equation $(x lo g x) \frac{d y}{d x} + y = 2 x lo g x, (x \geq 1)$ . Then $y (e)$ is equal to:

Solution of the differential equation $cos x d y = y (sin x - y) d x, 0 < x < \frac{π}{2}$ is

Let $f (x)$ be a function satisfying $f^{''} (x) = f (x)$ with $f (0) = 1$ and $g (x)$ be a function that satisfies $f (x) + g (x) = x^{2}$ . Then the value of the integral $\int_{0}^{1} f (x) g (x) d x$ , is

The solution of the differential equation $(1 + y^{2}) + (x - e^{t a n^{- 1} y}) \frac{d y}{d x} = 0$ , is

Let $f : (0, \infty) \to R$ be a differentiable function such that $f^{'} (x) = 2 - \frac{f ( x )}{x}$ for all $x \in (0, \infty)$ and $f (1) \neq = 1$ . Then

A solution curve of the differential equation $(x^{2} + x y + 4 x + 2 y + 4) \frac{d y}{d x} - y^{2} = 0, x > 0$ , passes through the point (1, 3). Then the solution curve

Let $y (x)$ be a solution of the differential equation $(1 + e^{x}) y^{'} + y e^{x} = 1$ . If $y (0) = 2$ , then which of the following statement is (are) true?

The function $y = f (x)$ is the solution of the differential equation $\frac{d y}{d x} + \frac{x y}{x ^{2} - 1} = \frac{x ^{4} + 2 x}{1 - x ^{2}}$ in $(- 1, 1)$ satisfying $f (0) = 0$ . Then $\int_{- 3 /2}^{3 /2} f (x) d x$ is

If $y (x)$ satisfies the differential equation $y^{'} - y tan x = 2 x sec x$ and $y (0) = 0$ , then

Let $f (x)$ be differentiable on the interval $(0, \infty)$ such that $f (1) = 1$ , and $lim_{t \to x} \frac{t ^{2} f ( x ) - x ^{2} f ( t )}{t - x} = 1$ for each $x > 0$ . Then $f (x)$ is

For the primitive integral equation $y d x + y^{2} d y = x d y$ ; $x \in R, y > 0, y (1) = 1$ , then $y (- 3)$ is

If $y (t)$ is a solution of $(1 + t) \frac{d y}{d t} - t y = 1$ and $y (0) = - 1$ , then $y (1)$ is equal to

A solution of the differential equation $(\frac{d y}{d x})^{2} - x \frac{d y}{d x} + y = 0$ is

Let $f : [1, \infty) \to [2, \infty)$ be a differentiable function such that $f (1) = 2$ . If $6 \int_{1}^{x} f (t) d t = 3 x f (x) - x^{3}$ for all $x \geq 1$ , then the value of $f (2)$ is .........

Let $f : R \to R$ be a continuous function which satisfies $f (x) = \int_{0}^{x} f (t) d t$ . Then the value of $f (ln 5)$ is

Let $f (x), x \geq 0$ , be a non-negative continuous function, and let $F (x) = \int_{0}^{x} f (t) d t, x \geq 0$ . If for some $c > 0, f (x) \leq c F (x)$ for all $x \geq 0$ , then show that $f (x) = 0$ for all $x \geq 0$ .

Let $y = y (x)$ be the solution of the differential equation $sin x \frac{d y}{d x} + y cos x = 4 x, x \in (0, π)$ . If $y (π /2) = 0$ , then $y (π /6)$ is equal to :

Let $y = y (x)$ be the solution of the differential equation $\frac{d y}{d x} + 2 y = f (x)$ , where $f (x) = {1, 0, x \in [0, 1] otherwise$ . If $y (0) = 0$ , then $y (3/2)$ is :

If $y = y (x)$ is the solution of the differential equation $x \frac{d y}{d x} + 2 y = x^{2}$ satisfying $y (1) = 1$ , then $y (1/2)$ is equal to :