Differential Equations(12th Grade > Mathematics ) Questions and answers for exam Preparation

12th Grade > Mathematics

DIFFERENTIAL EQUATIONS MCQs

Total Questions : 60 | Page 2 of 6 pages

Question 11. The orthogonal trajectories of the family of curves

a^{n - 1} y = x^{n}

are given by

xn+n2y= constant
ny2+x2= constant
n2x+yn= constant
n2x−yn= constant

Discuss Question

Answer: Option B. -> ny2+x2= constant
:
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}

- \frac{d x}{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 12. The order of the differential equation whose general solution is given by

y = C_{1} e^{2 x + C_{2}} + C_{3} e^{x} + C_{4} s i n (x + C_{5})

Discuss Question

Answer: Option B. -> 4
:
B

y = C_{1} e^{2 x + C_{2}} + C_{3} e^{x} + C_{4} s i n (x + C_{5}) = C_{1} . e^{C_{2}} e^{2 x} + C_{3} e^{x} + C_{4} (s i n x c o s C_{5} + c o s x s i n C_{5}) = A e^{2 x} + C_{3} e^{x} + B s i n x + D c o s x

Here,

A = C_{1} e^{C_{2}}, B = C_{4} c o s C_{5}, D = C_{4} s i n C_{5}

(Since equation consists of four arbitrary constants)

∴

order of differential equation = 4.

Question 13. The solution of the differential equation

(x^{2} s i n^{3} y - y^{2} c o s x) d x + (x^{3} c o s y s i n^{2} y - 2 y s i n x) d y = 0

x3sin3y=3y2sinx+C
x3sin3y+3y2sinx=C
x2sin3y+y3sinx=C
2x2siny+y2sinx=C

Discuss Question

Answer: Option A. -> x3sin3y=3y2sinx+C
:
A

(x^{2} s i n^{3} y - y^{2} c o s x) d x + (x^{3} c o s y s i n^{2} y - 2 y s i n x) d y = 0 \frac{d y}{d x} = \frac{y^{2} c o s x - x^{2} s i n^{3} y}{x^{3} c o s y s i n^{2} - 2 y s i n x} (x^{3} c o s y s i n^{2} y - 2 y s i n x) d y = (y^{2} c o s x - x^{2} s i n^{3} y) d x = 0 (\frac{x^{3}}{3} d s i n^{3} y - s i n d y^{2}) - s i n^{3} y d (\frac{x^{3}}{3}) + y^{2} d s i n x = 0

\frac{x^{3}}{3} d s i n^{2} y + s i n^{3} y d (\frac{x^{3}}{3}) - (s i n d y^{2} + y^{2} d s i n x)

d (\frac{x^{3}}{3} s i n^{3} y) - d (y^{2} s i n x) = 0 \frac{x^{3}}{3} s i n^{3} y - y^{2} s i n x = c

Question 14. If

x \frac{d y}{d x} = y (l o g y - l o g x + 1)

, then the solution of the equation is

y log(xy)=cx
x log(yx)=cy
log(yx)=cx
log(xy)=cx

Discuss Question

Answer: Option C. -> log(yx)=cx
:
C

\frac{d y}{d x} = \frac{y}{x} (l o g \frac{y}{x} + 1)

Put

y = v x \Rightarrow \frac{d y}{d x} = v + x \frac{d v}{d x}

∴ v + x \frac{d v}{d x} = v l o g v + v \Rightarrow \frac{1}{v l o g v} d v = \frac{1}{x} d x \Rightarrow \int \frac{1}{v l o g v} d v = \int \frac{1}{x} d x \Rightarrow l o g (l o g v) = l o g x + l o g c

\Rightarrow l o g \frac{y}{x} = c x

Question 15. The degree and order of the differential equation of the family of all parabolas whose axis is x–axis, are respectively

Discuss Question

Answer: Option A. -> 1,2
:
A
Equation of family of parabolas with x-axis as axis is

y^{2} = 4 a (x + α)

where

a, α

are two arbitrary constants. So differential equation is of order 2 and degree 1.

Question 16. The solution of

\frac{d y}{d x} = \frac{y}{x} + t a n \frac{y}{x}

y sin(yx)=cx
y sin(yx)=cy
sin(xy)=cx
sin(yx)=cx

Discuss Question

Answer: Option D. -> sin(yx)=cx
:
D
Put

y = v x . T h e n \frac{d y}{d x} = v + x \frac{d v}{d x}

Given equation is

\frac{d y}{d x} = \frac{v}{x} + t a n \frac{y}{x} \Rightarrow v + x t a n v \Rightarrow c o t v d v = \frac{d x}{x} \Rightarrow l o g s i n v = l o g x + l o g c

\Rightarrow s i n v = c x \Rightarrow s i n (\frac{y}{x} = c x)

Question 17. If

y_{1} (x)

is a solution of the differential equation

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

, then a solution of the differential equation

\frac{d y}{d x} + f (x) y = r (x)

1y1(x)∫r(x)y1(x)dx
y1(x)∫r(x)y1(x)dx
∫r(x)y1(x)dx
None of these

Discuss Question

Answer: Option A. -> 1y1(x)∫r(x)y1(x)dx
:
A

\frac{d y}{d x} - f (x) . y = 0 \frac{d y}{y} = f (x) d x

y = \int f (x) d x

y_{1} (x) = e^{\int f (x) d x}

Then for given equation I.F =

e^{\int f (x) d x}

Hence Solution

y . y_{1} (x) = \int r (x) . y_{1} (x) d x

y = \frac{1}{y_{1} (x)} \int r (x) . y_{1} (x) d x

Question 18. Solution of the equation

x d y = (y + x \frac{f (\frac{y}{x})}{f^{'} (\frac{y}{x})}) d x

f(xy)=cy
f(yx)=cx
f(yx)=cxy
f(yx)=0

Discuss Question

Answer: Option B. -> f(yx)=cx
:
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 19. 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}})

y =k
y =kx
y=k(ex+1)
y=k(ex−1)

Discuss Question

Answer: Option D. -> y=k(ex−1)
:
D

\frac{d p}{d x} = P (where p = \frac{d y}{d x})

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 20. 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

Discuss Question

Answer: Option A. -> TTT
:
A