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

12th Grade > Mathematics

DIFFERENTIAL EQUATIONS MCQs

Total Questions : 60 | Page 3 of 6 pages

Question 21. The general solution of the differential equation

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

tan x = (c + sec x)y
sec y = (c + tan y )x
sec x = (c + tan x)y
None of these

Discuss Question

Answer: Option C. -> sec x = (c + tan x)y
:
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 22. 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 23. 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

2y=x2−x
y=x2−x
y=x−x2
y=2(x−x2)

Discuss Question

Answer: Option C. -> y=x−x2
:
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 24. 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 25. 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 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 26. 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 27. 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})}

Question 28. 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 29. 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 30. A function y =f(x) has a second order derivative f"=6(x-1). If its graph passes through the point(2,1) and at that point the tangent to the graph is y =3x -5, then the function is