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

12th Grade > Mathematics

DIFFERENTIAL EQUATIONS MCQs

Total Questions : 60 | Page 4 of 6 pages

Question 31. The solution lof

\frac{d y}{d x} - x t a n (y - x) = 1

c eex22=sin(y−x)
c eex22=sin(y+x)
c eex22=sin(y−x)2
c eex22=cos(y−x)

Discuss Question

Answer: Option A. -> c eex22=sin(y−x)
:
A
Put y-x = z. Then

\frac{d y}{d x} - 1 = \frac{d z}{d x} \Rightarrow \frac{d y}{d x} = 1 + \frac{d z}{d x}

Given

\frac{d y}{d x} - x t a n (y - x) = 1 \Rightarrow 1 + \frac{d z}{d x} - x t a n z = 1 \Rightarrow \frac{d z}{d x} = x t a n z \Rightarrow \frac{1}{t a n z} d z = x d x \Rightarrow \int c o t z d z = \int x d x

\Rightarrow l o g | s i n z | - l o g c = \frac{x^{2}}{2} \Rightarrow l o g | \frac{s i n z}{c} | = \frac{x^{2}}{2} = \frac{s i n z}{c} = \frac{e^{x^{2}}}{2} \Rightarrow s i n (y - x) = c \frac{e^{x^{2}}}{2}

∴

The solution is

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

, where c is arbitrary constant.

Question 32. The order of the differential equation of all tangent lines to the parabola

y = x^{2}

Discuss Question

Answer: Option A. -> 1
:
A
The parametric form of the given equation is

x = t, y = t^{2}

.The equation of any tangent at t is

2 x t = y + t^{2}

, Differentiating we get

2 t = y_{1} (= \frac{d y}{d x})

putting this value in the above equation, we have

2 x \frac{y_{1}}{2} = y + {(\frac{y_{1}}{2})}^{2} \Rightarrow 4 x y_{1} = 4 y + y_{1}^{2}

The order of this equation is 1
Hence (A) is the correct answer

Question 33. The degree of the differential equation satisfying the relation

\sqrt{1 + x^{2}} + \sqrt{1 + y^{2}} = λ (x \sqrt{1 + y^{2}} - y \sqrt{1 + x^{2}})

1
2
3
none of these

Discuss Question

Answer: Option A. -> 1
:
A
On Putting

x = t a n A, y = t a n B

we get

s e c A + s e c B = λ (t a n A s e c B - t a n B s e c A) c o s A + c o s B = λ (s i n A - s i n B) t a n (\frac{A - B}{2}) = \frac{1}{λ} t a n^{- 1} x - t a n^{- 1} y = 2 t a n^{- 1} \frac{1}{λ}

On differentiating

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

Question 34. The general solution of

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

is
[EAMCET 2003]

tan−1(xy)+log y+c=0
2 tan−1(xy)+log x+c=0
log(y+√x2+y2)+log y+c=0
sin h−1(xy)+log y+c=0

Discuss Question

Answer: Option A. -> tan−1(xy)+log y+c=0
:
A

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

Put

v = x / y \Rightarrow x = v y \Rightarrow \frac{d x}{d y} = v + y \frac{d v}{d y} v + y \frac{d v}{d y} + v^{2} - v + 1 = 0 \Rightarrow \frac{d v}{v^{2} + 1} + \frac{d y}{y} = 0 \Rightarrow \int \frac{d v}{v^{2} + 1} + \int \frac{d y}{y} = 0 \Rightarrow t a n^{- 1} (v) + l o g y + C = 0 \Rightarrow t a n^{- 1} (x / y) + l o g y + c = 0.

Question 35. Solution of the differential equation :

\frac{d y}{d x} = \frac{3 x^{2} y^{4} + 2 x y}{x^{2} - 2 x^{3} y^{3}}

x2y2+x2y=c
x3y2+x2y=c
x3y2+y2x=c
x2y3+x2y=c

Discuss Question

Answer: Option B. -> x3y2+x2y=c
:
B

x^{2} d y - 2 x^{3} y^{3} d y = 3 x^{2} y^{4} d x + 2 x y d x \Rightarrow x^{2} d y - 2 x y d x = 3 x^{2} y^{4} d x + 2 x^{3} y^{3} d y \Rightarrow \frac{2 x y d x - x^{2} d y}{y^{2}} + 3 x^{2} y^{2} d x + 2 x^{3} y d y = 0 \Rightarrow d (\frac{x^{2}}{y}) + d (x^{3} y^{2}) = 0 \Rightarrow \frac{x^{2}}{y} + x^{3} y^{2} = C

Question 36. 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 37. The integrating factor of the differential equation

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

is
[MP PET 1995; Pb. CET 2002]

tan x
secx
-sec x
cot x

Discuss Question

Answer: Option B. -> secx
:
B
The differential equation
is

\frac{d y}{d x} - y t a n x = - y^{2} s e c x I . F . = e^{- \int t a n x d x}

This is Bernoulli's equation i.e. reducible to
linear equation.
Dividing the equation by

y^{2},

we get

\frac{1}{y^{2}} \frac{d y}{d x} - \frac{1}{y} t a n x = - s e c x . . . . . . . . . . . . (i)

Put

\frac{1}{y} = y \Rightarrow - \frac{1}{y^{2}} \frac{d y}{d x} = \frac{d Y}{d x}

Equation (i) reduces to

- \frac{d y}{d x} = - y t a n x = - s e c x \Rightarrow \frac{d Y}{d x} + Y t a n x = s e c x,

Which is a linear equation
Hence

I . F . = e^{- \int t a n x d x} = s e c x .

Question 38. 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