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

12th Grade > Mathematics

DIFFERENTIAL EQUATIONS MCQs

Total Questions : 60 | Page 1 of 6 pages

Question 1. 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 2. 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 3. 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 4. The solution of

\frac{d y}{d x} + \frac{x}{y} = x^{2}

1y=cx−xlog x
1x=cy−ylog y
1x=cx−xlog y
1y=cx−ylog x

Discuss Question

Answer: Option B. -> 1x=cy−ylog y
:
B

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

Put

x^{- 1} = t . T h e n - x^{- 2} \frac{d x}{d y} = \frac{d t}{d y} \Rightarrow x^{- 2} \frac{d x}{d y} = - \frac{d t}{d y} ∴ - \frac{d t}{d y} + \frac{t}{y} = 1 \Rightarrow \frac{d t}{d y} - \frac{1}{y} t = - 1

It is linear in t. Here

P = - \frac{1}{y} . Q = - 1

I . F = e^{\int (\frac{1}{y}) d y} = e^{- l o g y} = \frac{1}{y}

∴

Solution is

t (\frac{1}{y}) = \int (- 1) \frac{1}{y} d y = - l o g y + c \Rightarrow t = - y l o g y + c y \Rightarrow x^{- 1} c y - y l o g y .

Question 5. 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 6. 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

(x−1)2
(x−1)3
(x+1)2
(x+1)3

Discuss Question

Answer: Option B. -> (x−1)3
:
B
Since f"(x)=6(x-1)

\Rightarrow f^{'} (x) = 3 (x - 1)^{2} + c

(integrating) ----(i)
Also, at the point (2,1), the tangent to the graph is y =3x-5and slope of thetangent = 3

\Rightarrow f^{'} (2) = 3

3 (2 - 1)^{3} + c = 3

[from eq(i)

\Rightarrow 3 + c = 3 \Rightarrow c = 0

From Eq (i) we have

f^{'} (x) = 3 (x - 1)^{2}

\Rightarrow f (x) = (x - 1)^{3} + k

(Integrating)----(ii)

∴ 1 = (2 - 1)^{3} + k \Rightarrow k = 0

Hence the equation of the function is

f (x) = (x - 1)^{3}

Question 7. 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 8. The curves satisfying the differential equation

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

are

ellipses and hyperbolas
ellipses and parabola
ellipses and straight lines
circles and ellipses

Discuss Question

Answer: Option A. -> ellipses and hyperbolas
:
A
The given equation is linear DE and can be written as

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

Its integrating factor is

e^{\int \frac{x}{1 - x^{2}} d x} = e^{- (\frac{1}{2}) l o g (1 - x^{2})} = \frac{1}{\sqrt{1 - x^{2}}} - 1 < x < 1

and

i f x^{2} > 1

then I.F.=

\frac{1}{\sqrt{x^{2} - 1}}

\frac{d}{d x} (y \frac{1}{\sqrt{1 - x^{2}}}) = \frac{a x}{(1 - x^{2})^{\frac{3}{2}}} = - \frac{1}{2} a \frac{- 2 x}{(1 - x^{2})^{\frac{3}{2}}}

\Rightarrow y \frac{1}{\sqrt{1 - x^{2}}} = \frac{a}{\sqrt{1 - x^{2}}} + C \Rightarrow y = a + C \sqrt{1 - x^{2}}

\Rightarrow (y - a)^{2} = C^{2} (1 - x^{2}) \Rightarrow (y - a)^{2} + C^{2} x^{2} = C^{2}

Thus if -1 < x < 1 the given equation represents an ellipse. If

x^{2} > 1

then the solution is of the form

- (y - a)^{2} + C^{2} x^{2} = C^{2}

which represents a hyperbola.

Question 9. 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 10. The solution of

\frac{d y}{d x} + \frac{x}{y} = x^{2}

1y=cx−xlog x
1x=cy−ylog y
1x=cx−xlog y
1y=cx−ylog x

Discuss Question

Answer: Option B. -> 1x=cy−ylog y
:
B

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

Put

x^{- 1} = t . T h e n - x^{- 2} \frac{d x}{d y} = \frac{d t}{d y} \Rightarrow x^{- 2} \frac{d x}{d y} = - \frac{d t}{d y} ∴ - \frac{d t}{d y} + \frac{t}{y} = 1 \Rightarrow \frac{d t}{d y} - \frac{1}{y} t = - 1

It is linear in t. Here

P = - \frac{1}{y} . Q = - 1

I . F = e^{\int (\frac{1}{y}) d y} = e^{- l o g y} = \frac{1}{y}

∴

Solution is

t (\frac{1}{y}) = \int (- 1) \frac{1}{y} d y = - l o g y + c \Rightarrow t = - y l o g y + c y \Rightarrow x^{- 1} c y - y l o g y .

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