The solution of the differential equation d y d x + y 2 sec ⁡ x = tan ⁡ x 2 y , where 0 ≤ x < π 2 and y 0 = 1 , is given by

y 2 = 1 - x sec ⁡ x + tan ⁡ x

y 2 = 1 + x sec ⁡ x + tan ⁡ x

y = 1 + x sec ⁡ x + tan ⁡ x

y = 1 - x sec ⁡ x + tan ⁡ x

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

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To reduce the differential equation $\frac{d y}{d x} + P (x) . y = Q (x) . y^{n}$ to the linear form, the substitution is:

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Important Questions on Differential Equation

MEDIUM

Mathematics>Integral Calculus>Differential Equation>Linear Differential Equations

\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

HARD

Mathematics>Integral Calculus>Differential Equation>Linear Differential Equations

The solution of the differential equation

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

, where

0 \leq x < \frac{π}{2}

and

y (0) = 1

, is given by

HARD

Mathematics>Integral Calculus>Differential Equation>Linear Differential Equations

f (x)

is a differentiable function in 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 (\frac{3}{2})

is equal to

HARD

Mathematics>Integral Calculus>Differential Equation>Linear Differential Equations

Let

y = y (x)

be the solution of the differential equation,

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

such that

y (0) = 0 .

\sqrt{a} y (1) = \frac{π}{32},

then the value of

a

HARD

Mathematics>Integral Calculus>Differential Equation>Linear Differential Equations

If a curve passes through the point

(1, - 2)

and has slope of the tangent at any point

(x, y)

on it as

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

, then the curve also passes through the point

MEDIUM

Mathematics>Integral Calculus>Differential Equation>Linear Differential Equations

c o s x \frac{d y}{d x} - y s i n x = 6 x

(0 < x < \frac{π}{2})

and

y (\frac{π}{3}) = 0,

then

y (\frac{π}{6})

is equal to

HARD

Mathematics>Integral Calculus>Differential Equation>Linear Differential Equations

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}{\sqrt{1 - x^{2}}}

(- 1, 1)

satisfying

f (0) = 0 .

Then

\int_{- \frac{\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}} f (x) d x

MEDIUM

Mathematics>Integral Calculus>Differential Equation>Linear Differential Equations

The solution of the differential equation

\frac{d θ}{d t} = - k (θ - θ_{0})

, where

k

is a constant, is ……

HARD

Mathematics>Integral Calculus>Differential Equation>Linear Differential Equations

Let

f (x)

be a differentiable function such that

f^{'} (x) = 7 - \frac{3}{4} \frac{f (x)}{x}, (x > 0)

and

f (1) \neq 4 .

Then

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

HARD

Mathematics>Integral Calculus>Differential Equation>Linear Differential Equations

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

and

y (0) = 1

, then

y (π)

is equal to

HARD

Mathematics>Integral Calculus>Differential Equation>Linear Differential Equations

y = y (x)

is the solution of the differential equation

\frac{d y}{d x} = (t a n x - y) {s e c}^{2} x

x \in (- \frac{π}{2}, \frac{π}{2})

, such that

y (0) = 0

, then

y (- \frac{π}{4})

is equal to:

EASY

Mathematics>Integral Calculus>Differential Equation>Linear Differential Equations

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

HARD

Mathematics>Integral Calculus>Differential Equation>Linear Differential Equations

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 Equation>Linear Differential Equations

The solution of the differential equation

x \frac{d y}{d x} + 2 y = x^{2}, (x \neq 0)

with

y (1) = 1

, is

HARD

Mathematics>Integral Calculus>Differential Equation>Linear Differential Equations

Consider the differential equation,

y^{2} d x + (x - \frac{1}{y}) d y = 0

. If value of

y

1

when

x = 1

, then the value of

x

for which

y = 2,

MEDIUM

Mathematics>Integral Calculus>Differential Equation>Linear Differential Equations

The solution of

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

is :

MEDIUM

Mathematics>Integral Calculus>Differential Equation>Linear Differential Equations

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

MEDIUM

Mathematics>Integral Calculus>Differential Equation>Linear Differential Equations

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

HARD

Mathematics>Integral Calculus>Differential Equation>Linear Differential Equations

The curve satisfying the differential equation,

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

and passing through the point

(1, 1)

also passes through the point

MEDIUM

Mathematics>Integral Calculus>Differential Equation>Linear Differential Equations

Let

y (x)

be the solution of the differential equation

(x \log x) \frac{d y}{d x} + y = 2 x \log x, (x \geq 1)

. Then

y (e)

is equal to

To reduce the differential equation dydx+Px.y=Qx.yn to the linear form, the substitution is:

Important Questions on Differential Equation

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To reduce the differential equation $\frac{d y}{d x} + P (x) . y = Q (x) . y^{n}$ to the linear form, the substitution is: