Free Differential Equations 03 Practice Test - 12th Grade - Commerce

Question 1

The general solution of the differential equation $\frac{d y}{d x} = y t a n x - y^{2} s e c x$ is

A. tan x = (c + sec x)y

B. sec y = (c + tan y )x

C. sec x = (c + tan x)y

D. None of these

SOLUTION

Solution : C

We have $\frac{d y}{d x} = y t a n x - y^{2} s e c x \Rightarrow \frac{1}{y^{2}} \frac{d y}{d x} - \frac{1}{y} t a n x = - s e c x$
Putting $\frac{1}{y} = v \Rightarrow \frac{- 1}{y^{2}} \frac{d y}{d x} = \frac{d v}{d x}$ , we obtain
$\frac{d v}{d x} + t a n x . v = s e c x$ which is linear
$I . F = e^{\int t a n x d x} = e^{l o g s e c x} = s e c x$
$∴$ The solution is
$v s e c x = \int s e c^{2} x d x + c \Rightarrow \frac{1}{y} s e c x = t a n x + c$
$\Rightarrow s e c x = y (c + t a n x)$

Question 2

If integrating factor of $x (1 - x^{2}) d y + (2 x^{2} y - y - a x^{3}) d x = 0$ is $e^{\int P d x}$ , then P is equal to

\frac{2 x^{2} - a x^{3}}{x (1 - x^{2})}

(2 x^{2} - 1)

\frac{2 x^{2} - 1}{a x^{3}}

\frac{(2 x^{2} - 1)}{x (1 - x^{2})}

SOLUTION

Solution : D

$x (1 - x^{2}) d y + (2 x^{2} y - y - a x^{3}) d x = 0 \frac{d y}{d x} + \frac{(2 x^{2} - 1)}{x (1 - x^{2})} y = \frac{a x^{2}}{(1 - x^{2})}, ∴ P = \frac{2 x^{2} - 1}{x (1 - x^{2})}$ .

Question 3

Solution of the equation $x d y = (y + x \frac{f (\frac{y}{x})}{f^{'} (\frac{y}{x})}) d x$

f (\frac{x}{y}) = c y

f (\frac{y}{x}) = c x

f (\frac{y}{x}) = c x y

f (\frac{y}{x}) = 0

SOLUTION

Solution : B

We have, $x d y = (y + \frac{x f (\frac{y}{x})}{f^{'} (\frac{y}{x})}) d x$
$\Rightarrow \frac{d y}{d x} = \frac{y}{x} + \frac{f (\frac{y}{x})}{f^{'} (\frac{y}{x})}$ which is homogenous
Putting $y = v x \Rightarrow \frac{d y}{d x} = v + x \frac{d v}{d x}$ , we obtain
$v + x \frac{d v}{d x} = v + \frac{f (v)}{f^{'} (v)} \Rightarrow \frac{f^{'} (v)}{f (v)} d v = \frac{d x}{x}$
Integrating, we get log $f (v) = l o g x + l o g c$
$\Rightarrow l o g f (v) = l o g c x \Rightarrow f (\frac{y}{x}) = c x$

Question 4

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

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

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

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

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

SOLUTION

Solution : A

$\frac{d x}{d y} + \frac{1}{1 + y^{2}} x = \frac{1}{1 + y^{2}} e^{t a n^{- 1} y}$
$I . F = e^{\int \frac{1}{1 + y^{2}} d y} = e^{t a n^{- 1} y}$
$∴$ Solution is $x . e^{t a n^{- 1}} y$
$= \int e^{t a n^{- 1} y} . \frac{1}{1 + y^{2}} e^{t a n^{- 1}} d y = \frac{1}{2} e^{2 t a n^{- 1} y} + \frac{1}{2} k \Rightarrow 2 x e^{t a n^{- 1} y} = e^{2 t a n^{- 1} y} + k$

Question 5

The orthogonal trajectories of the family of curves $a^{n - 1} y = x^{n}$ are given by

x^{n} + n^{2} y

=constant

n y^{2} + x^{2}

=constant

n^{2} x + y^{n}

=constant

n^{2} x - y^{n}

=constant

SOLUTION

Solution : B

Differentiating, we have $a^{n - 1} \frac{d y}{d x} = n x^{n - 1} \Rightarrow a^{n - 1} = n x^{n - 1} \frac{d x}{d y}$
Putting this value in the given equation, we have $n x^{n - 1} \frac{d x}{d y} y = x^{n}$
Replacing $\frac{d y}{d x}$ by $- \frac{d y}{d y}$ , we have $n y = - x \frac{d x}{d y}$
$\Rightarrow n y d y + x d x = 0 \Rightarrow n y^{2} + x^{2}$ =constant. Which is the required family of orthogonal trajectories.

Question 6

The solution of $\frac{d y}{d x} + 1 = e^{x + y}$ is

e^{- (x + y)} + x + c = 0

e^{- (x + y)} - x + c = 0

e^{x + y} + x + c = 0

e^{x + y} - x + c = 0

SOLUTION

Solution : A

$x + y = z \Rightarrow 1 + \frac{d y}{d x} = \frac{d z}{d x} \frac{d y}{d x} + 1 = e^{x + y} \Rightarrow \frac{d y}{d x} = e^{- z} d z = d x \Rightarrow \int e^{- z} d z = \int d x \Rightarrow - e^{- z} = x + c \Rightarrow e^{- (x + y)} + x + c = 0.$

Question 7

Solution to the differential equation $\frac{x + \frac{x^{3}}{3!} + \frac{x^{5}}{5!} + . . . . .}{1 + \frac{x^{2}}{2!} + \frac{x^{4}}{4!} + . . . . .} = \frac{d x - d y}{d x + d y}$ is

2 y e^{2 x} = C . e^{2 x} + 1

2 y e^{2 x} = C . e^{2 x} - 1

y e^{2 x} = C . e^{2 x} + 2

2 x e^{2 y} = C . e^{x} - 1

SOLUTION

Solution : B

Applying C and D, we get
$\frac{d y}{d x} = \frac{e^{- x}}{e^{x}} = e^{- 2 x} \Rightarrow 2 y = - e^{- 2 x} + C$
or $2 y e^{2 x} = C . e^{2 x} - 1$ .

Question 8

A curve is such that the mid point of the portion of the tangent intercepted between the point where the tangent is drawn and the point where the tangent meets y-axis, lies on the line y = x. If the curve passes through (1, 0), then the curve is

2 y = x^{2} - x

y = x^{2} - x

y = x - x^{2}

y = 2 (x - x^{2})

SOLUTION

Solution : C

The point on y-axis is $(0, y - x \frac{d y}{d x})$
According to given condition,
$\frac{x}{2} = y - \frac{x}{2} \frac{d y}{d x} \Rightarrow \frac{d y}{d x} = 2 \frac{y}{x} - 1$
Putting $\frac{y}{x} = v$ we get
$x \frac{d v}{d x} = v - 1 \Rightarrow l n ∣ ∣ \frac{y}{x} - 1 ∣ ∣ = l n | x | + c \Rightarrow 1 - \frac{y}{x} = x$ (as f(1)=0).

Question 9

The family of curves passing through (0,0) and satisfying the differential equation $\frac{y_{2}}{y_{1}} = 1$ (where $y_{n} = \frac{d^{n} y}{d x^{n}})$ is

A. y =k

B. y =kx

y = k (e^{x} + 1)

y = k (e^{x} - 1)

SOLUTION

Solution : D

$\frac{d p}{d x} = P (where p = \frac{d y}{d x})$
ln $P = x + c \Rightarrow p = e^{x + c}$
$\frac{d y}{d x} = k e^{x} y = k e^{x} + λ$
Satisfying (0,0), So $λ = - k$
$y = k (e^{x} - 1)$

Question 10

S1: The differential equation of parabolas having their vertices at the origin and foci on the x-axis is an equation whose variables are separable
S2: The differential equation of the straight lines which are at a fixed distance p from the origin is an equation of degree 2
S3: The differential equation of all conics whose both axes coincide with the axes of coordinates is an equation of order 2

A. TTT

B. TFT

C. FFT

D. TTF

SOLUTION

Solution : A

$S_{1}$ -Equation of parabola is $y^{2} = \pm 4 a x$
$2 y \frac{d y}{d x} = \pm r a$
D.E of parabola $\Rightarrow y^{2} = 2 y x \frac{d y}{d x}$
$2 \frac{d y}{y} = \frac{d x}{x}$
Which is variable seperable
$S_{2}$ -Equation of line which is fixed distance. P from origin can be equation of tangent to circle
$s^{2} + y^{2} = p^{2}$
Line is $y = m x + p \sqrt{1 + m^{2}} (m = \frac{d y}{d x})$
${(y - x \frac{d y}{d x})}^{2} = P (1 + {(\frac{d y}{d x})}^{2})$
So, degree is 2
$S_{3}$ -Equation of conic whose both axis co-incide with co-ordinate axis is $a x^{2} + b y^{2} = 1$
As there are two constants, so order of D.E is 2

Free Differential Equations 03 Practice Test - 12th Grade - Commerce

Question 1

SOLUTION

Question 2

SOLUTION

Question 3

SOLUTION

Question 4

SOLUTION

Question 5

SOLUTION

Question 6

SOLUTION

Question 7

SOLUTION

Question 8

SOLUTION

Question 9

SOLUTION

Question 10

SOLUTION

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