The differential equation d y d x = x 1 - x 2 , determine a family of semi-circles with

Fixed radius 1 and variable centres along the y -axis

Variable radii and a fixed centre at 1 , 0

Variable radii and a fixed centre at - 1 , 0

Fixed radius 1 and variable centres along the x -axis

The differential equation d y d x = x 1 - x 2 , determine a family of semi-circles with

Fixed radius 1 and variable centres along the y -axis

The solution curve of the differential equation, 1 + e - x 1 + y 2 d y d x = y 2 which passes through the point 0 , 1 , is

y 2 + 1 = y log e 1 + e - x 2 + 2

y 2 + 1 = y log e 1 + e x 2 + 2

The general solution of the differential equation 1 + x 2 + y 2 + x 2 y 2 + x y d y d x = 0 (where C is a constant of integration)

1 + y 2 + 1 + x 2 = 1 2 log e 1 + x 2 + 1 1 + x 2 - 1 + C

1 + y 2 + 1 + x 2 = 1 2 log e 1 + x 2 - 1 1 + x 2 + 1 + C

1 + y 2 - 1 + x 2 = 1 2 log e 1 + x 2 - 1 1 + x 2 + 1 + C

1 + y 2 - 1 + x 2 = 1 2 log e 1 + x 2 + 1 1 + x 2 - 1 + C

The law of motion of a body moving along a straight line is x = 1 2 v t . x being its distance from a fixed point on the line at time t and v is its velocity there. Then

acceleration f varies directly with x

acceleration f varies inversely with x

acceleration f varies directly with t

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

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Important Questions on Differential Equations

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

Let $f : (0, \infty) \to (0, \infty)$ be a differentiable function such that $f (1) = e$ and $\lim_{t \to x} \frac{t^{2} f^{2} (x) - x^{2} f^{2} (t)}{t - x} = 0 .$ If $f (x) = 1$ , then $x$ is equal to:

HARD

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

The particular solution of the differential equation

\log (\frac{d y}{d x}) = x

, when

x = 0

y = 1

is …..

HARD

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

y = y (x)

is the solution of the differential equation

\frac{5 + e^{x}}{2 + y} \cdot \frac{d y}{d x} + e^{x} = 0

satisfying

y (0) = 1

then value of

y (\log_{e} 13)

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

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

HARD

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

The solution curve of the differential equation,

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

which passes through the point

(0, 1)

, 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{2 + \sin x}{y + 1} . \frac{d y}{d x} = - \cos x

y > 0, y (0) = 1

. If

y (π) = a

and

\frac{d y}{d x}

x = π

b

, then the ordered pair

(a, b)

is equal to

EASY

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

The solution of

y^{'} = e^{x - y} + x^{2} e^{- y}

HARD

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) = 1

and satisfying the equation

f (x + y) = f (x) f^{'} (y) + f^{'} (x) f (y)

for all

x, y \in R

. Then, the value of

\log_{e} (f (4))

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

Solution of the differential equation

\frac{d y}{d x} = \sin (x + y) + \cos (x + y)

is equal to

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

HARD

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

The solution of

\frac{d y}{d x} = \cos (x + y) + \sin (x + y)

, is

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

The general solution of the differential equation

\sqrt{1 + x^{2} + y^{2} + x^{2} y^{2}} + x y \frac{d y}{d x} = 0

(where C is a constant of integration)

HARD

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

f^{'} (x) = f (x) + \int_{0}^{1} f (x) d x, f (0) = 1

, then

f (x) =

EASY

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

The law of motion of a body moving along a straight line is

x = \frac{1}{2} v t

. x

being its distance from a fixed point on the line at time

t

and

v

is its velocity there. Then

MEDIUM

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

The general solution of the differential equation

\tan (y) d x + \sec^{2} (y) \cdot \tan (x) d y = 0

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

(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

The solution of

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

is:

The differential equation dydx=x1-x2, determine a family of semi-circles with

Important Questions on Differential Equations

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