The solution of the curves x d x + y d x x d y - y d x = 1 - x 2 - y 2 x 2 + y 2 are

circles of radius 1 2 through the origin

circles passing through the origin

The solution of the curves x d x + y d x x d y - y d x = 1 - x 2 - y 2 x 2 + y 2 are

circles of radius 1 2 through the origin

The solution of the differential equation d θ d t = - k θ - θ 0 , where k is a constant, is ……

θ = θ 0 + a e - k t where a is the constant of integration

θ = θ 0 + a e k t where a is the constant of integration

θ = 2 θ 0 - a e k t where a is the constant of integration

θ = 2 θ 0 - a e - k t where a is the constant of integration

Let y = y x be the solution of the differential equation, d y d x + y tan x = 2 x + x 2 tan x , x ∈ - π 2 , π 2 , such that y 0 = 1 . Then

y ' π 4 + y ' - π 4 = π 2 2 + 2

HARD

Earn 100

The solution of the curves $\frac{x d x + y d x}{x d y - y d x} = \sqrt{\frac{1 - x^{2} - y^{2}}{x^{2} + y^{2}}}$ are

50% studentsanswered this correctly

Important Questions on Differential Equations

MEDIUM

Mathematics>Integral Calculus>Differential Equations>Methods of Solving First Order, First Degree Differential Equation

If for

x \geq 0, y = y (x)

is the solution of the differential equation,

(x + y) d y = ({(x + 1)}^{2} + y - 3) d x, y (2) = 0

then

y (3)

is equal to ________

MEDIUM

Mathematics>Integral Calculus>Differential Equations>Methods of Solving First Order, First Degree Differential Equation

A tangent to the curve,

y = f (x)

P (x, y)

meets

x

-axis at

A

and

y

-axis at

B

. If

A P : B P = 1 : 3

and

f (1) = 1,

then the curve also passes through the point

MEDIUM

Mathematics>Integral Calculus>Differential Equations>Methods of Solving First Order, First Degree Differential Equation

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) .

y (\frac{π}{2}) = 0,

then

y (\frac{π}{6})

is equal to

MEDIUM

Mathematics>Integral Calculus>Differential Equations>Methods of Solving First Order, First Degree Differential Equation

The solution of the differential equation

\frac{d θ}{d t} = - k (θ - θ_{0})

, where

k

is a constant, is ……

EASY

Mathematics>Integral Calculus>Differential Equations>Methods of Solving First Order, First Degree Differential Equation

The solution of

\frac{d y}{d x} = 2^{y - x}

is:

HARD

Mathematics>Integral Calculus>Differential Equations>Methods of Solving First Order, First Degree Differential Equation

Let

y = y (x)

be the solution of the differential equation,

\frac{d y}{d x} + y \tan x = 2 x + x^{2} \tan x, x \in (- \frac{π}{2}, \frac{π}{2}),

such that

y (0) = 1 .

Then

HARD

Mathematics>Integral Calculus>Differential Equations>Methods of Solving First Order, First Degree Differential Equation

\frac{d y}{d x} + y \tan x = sin 2 x

and

y (0) = 1

, then

y (π)

is equal to

MEDIUM

Mathematics>Integral Calculus>Differential Equations>Methods of Solving First Order, First Degree Differential Equation

The solution of

\frac{d y}{d x} + y = e^{- x}, y (0) = 0

is :

HARD

Mathematics>Integral Calculus>Differential Equations>Methods of Solving First Order, First Degree Differential Equation

Let the population of rabbits surviving at a time

t

be governed by the differential equation

\frac{d p (t)}{d t} = \frac{1}{2} {p (t) - 400}

. If

p (0) = 100

, then

p (t)

equals

MEDIUM

Mathematics>Integral Calculus>Differential Equations>Methods of Solving First Order, First Degree Differential Equation

(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

EASY

Mathematics>Integral Calculus>Differential Equations>Methods of Solving First Order, First Degree Differential Equation

Let

y = y (x)

be the solution of the differential equation,

x \frac{d y}{d x} + y = x \log_{e} x, (x > 1)

. If

2 y (2) = \log_{e} 4 - 1

, then

y (e)

is equal to

MEDIUM

Mathematics>Integral Calculus>Differential Equations>Methods of Solving First Order, First Degree Differential Equation

Let

f : R \to R

be a differentiable function with

f (0) = 0

. If

y = f (x)

satisfies the differential equation

\frac{d y}{d x} = (2 + 5 y) (5 y - 2) .

If the value of

\lim_{x \to - \infty} f (x) = λ

, then

10 λ

is equal to

MEDIUM

Mathematics>Integral Calculus>Differential Equations>Methods of Solving First Order, First Degree Differential Equation

\frac{d y}{d x} = \frac{x y}{x^{2} + y^{2}}; y (1) = 1;

then a value of

x

satisfying

y (x) = e

is:

HARD

Mathematics>Integral Calculus>Differential Equations>Methods of Solving First Order, First Degree Differential Equation

f (x)

is a non-zero polynomial of degree four, having local extreme points at

x = - 1, 0, 1;

then the set

S = {x \in R : f (x) = f (0)}

contains exactly

MEDIUM

Mathematics>Integral Calculus>Differential Equations>Methods of Solving First Order, First Degree Differential Equation

Let

y = y (x)

be the solution curve of the differential equation,

(y^{2} - x) \frac{d y}{d x} = 1

, satisfying

y (0) = 1

. This curve intersects the

X -

axis at a point whose abscissa is

HARD

Mathematics>Integral Calculus>Differential Equations>Methods of Solving First Order, First Degree Differential Equation

The general solution of

\frac{dy}{dx} = \frac{2 x - y}{x + 2 y}

is given by

EASY

Mathematics>Integral Calculus>Differential Equations>Methods of Solving First Order, First Degree Differential Equation

The integrating factor of the differential equation

(2 x + 3 y^{2}) d y = y d x, (y > 0)

MEDIUM

Mathematics>Integral Calculus>Differential Equations>Methods of Solving First Order, First Degree Differential Equation

The particular solution of the differential equation

x d y + 2 y d x = 0

, when

x = 2 & y = 1

HARD

Mathematics>Integral Calculus>Differential Equations>Methods of Solving First Order, First Degree Differential Equation

y (x)

is the solution of the differential equation

(x + 2) \frac{d y}{d x} = x^{2} + 4 x - 9, x \neq - 2

and

y (0) = 0,

then

y (- 4)

is equal to

MEDIUM

Mathematics>Integral Calculus>Differential Equations>Methods of Solving First Order, First Degree Differential Equation

\frac{d y}{d x} + \frac{3}{\cos^{2} x} y = \frac{1}{\cos^{2} x}

x \in (- \frac{π}{3}, \frac{π}{3}),

and

y (\frac{π}{4}) = \frac{4}{3},

then

y (- \frac{π}{4})

equals

The solution of the curves xdx+ydxxdy-ydx=1-x2-y2x2+y2 are

Important Questions on Differential Equations

Learn and Prepare for any exam you want !

The solution of the curves $\frac{x d x + y d x}{x d y - y d x} = \sqrt{\frac{1 - x^{2} - y^{2}}{x^{2} + y^{2}}}$ are