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

12th Grade > Mathematics

DIFFERENTIAL EQUATIONS MCQs

Total Questions : 60 | Page 5 of 6 pages

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 42. 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}

2ye2x=C.e2x+1
2ye2x=C.e2x−1
ye2x=C.e2x+2
2xe2y=C.ex−1

Discuss Question

Answer: Option B. -> 2ye2x=C.e2x−1
:
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

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

Question 43. The family of curves passing through

(0, 0)

and satisfying the differential equation

\frac{y^{''}}{y^{'}} = 1

(where y^{'} = \frac{d y}{d x} and y^{''} = \frac{d^{2} y}{d x^{2}})

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})

ln p = x + c \Rightarrow p = e^{x + c}

\frac{d y}{d x} = k e^{x} \Rightarrow y = k e^{x} + λ

Satisfying

(0, 0)

, So

λ = - k

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

Question 44. 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 45. The solution of

\frac{d y}{d x} + 1 = e^{x + y}

e−(x+y)+x+c=0
e−(x+y)−x+c=0
ex+y+x+c=0
ex+y−x+c=0

Discuss Question

Answer: Option A. -> e−(x+y)+x+c=0
:
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 46. 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 47. 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 48. 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