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

12th Grade > Mathematics

DIFFERENTIAL EQUATIONS MCQs

Total Questions : 60 | Page 6 of 6 pages

Question 51. 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 52. 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 53. The solution of

(y (1 + x^{- 1}) + s i n y) d x + (x + l o g x + x c o s y) d y = 0

(1+y−1siny)+x−1logx=C
(y+siny)+xy log x=C
xy+ylogx+x sin y=C
None of these

Discuss Question

Answer: Option C. -> xy+ylogx+x sin y=C
:
C
The given equation can be written as

y (1 + x^{- 1}) d x + (x + l o g x) d y + s i n y d x + x c o s y d y = 0

\Rightarrow d (y (x + l o g x)) + d (x s i n y) = 0 \to y (x + l o g x) + x s i n y = C

Question 54. The solution of

y_{2} - 7 y_{1} + 12 y = 0

y=C1e3x+C2e4x
y=C1xe3x+C2e4x
y=C1e3x+C2xe4x
None of these

Discuss Question

Answer: Option A. -> y=C1e3x+C2e4x
:
A
The given equation can be written as

(\frac{d}{d x} - 3) (\frac{d y}{d x} - 4 y) = 0 . . . . (i)

\frac{d y}{d x} - 4 y = u

then (I) reduces to

\frac{d u}{d x} - 3 u = 0

\Rightarrow \frac{d u}{u} = 3 d x \Rightarrow u = C_{1} e^{3 x}

.Therefore, we have

\frac{d y}{d x} - 4 y = C_{1} e^{3 x}

which is a linear equation whose I.F.is

e^{- 4 x}

. So

\frac{d}{d x} (y e^{- 4 x}) = C_{1} e^{- x}

\Rightarrow y e^{- 4 x} = - C_{1} e^{- x} + C_{2} \Rightarrow y = C_{1} e^{3 x} + C_{2} e^{4 x}

Question 55. 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 56. The solution of the differential equation

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

, is

2x etan−1y,=e2tan−1y+k
x etan−1y,=etan−1y+k
x e2tan−1y,=e−tan−1y+k
(x−2)k etan−1y

Discuss Question

Answer: Option A. -> 2x etan−1y,=e2tan−1y+k
:
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 57. The order and degree of the differential equation

y = x \frac{d y}{d x} + \sqrt{a^{2} {(\frac{d y}{d x})}^{2} + b^{2}}

are

Discuss Question

Answer: Option A. -> 1,2
:
A
Given differential equation can be written as

y^{2} = x^{2} {(\frac{d y}{d x})}^{2} - 2 x y . \frac{d y}{d x} = a^{2} {(\frac{d y}{d x})}^{2} + b^{2} .

Hence it is of 1^storder and 2^nddegree differential equation.

Question 58. 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 59. The order and degree of the differential equation

{[4 + {(\frac{d y}{d x})}^{2}]}^{2 / 3} = \frac{d^{2} y}{d x^{2}}

are

Discuss Question

Answer: Option C. -> 2,3
:
C
Here power on the differential coefficient is fractional, therefore change it into positive integer, so

{[4 + {(\frac{d y}{d x})}^{2}]}^{2 / 3} = \frac{d^{2} y}{d x^{2}} \Rightarrow {[4 + {(\frac{d y}{d x})}^{2}]}^{2} = {[\frac{d^{2} y}{d x^{2}}]}^{3}

Hence order is 2 and degree is 3.

Question 60. If integrating factor of

x (1 - x^{2}) d y + (2 x^{2} y - y - a x^{3}) d x = 0

e^{\int P d x}

, then P is equal to

2x2−ax3x(1−x2)
(2x2−1)
2x2−1ax3
(2x2−1)x(1−x2)

Discuss Question

Answer: Option D. -> (2x2−1)x(1−x2)
:
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})}

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