Differential Equations Questions — JEE Mathematics

Q1

Let us consider a curve, y = f(x) passing through the point (

-

2, 2) and the slope of the tangent to the curve at any point (x, f(x)) is given by f(x) + xf'(x) = x2. Then :

A

{x^2} + 2xf(x) - 12 = 0

B

{x^3} + xf(x) + 12 = 0

C

{x^3} - 3xf(x) - 4 = 0

D

{x^2} + 2xf(x) + 4 = 0

Correct Answer

Option C

Solution

y + {{xdy} \over {dx}} = {x^2}

(given)

\Rightarrow {{dy} \over {dx}} + {y \over x} = x

If

{e^{\int {{1 \over x}dx} }} = x

Solution of DE

\Rightarrow y\,.\,x = \int {x\,.\,x\,dx}

\Rightarrow xy = {{{x^3}} \over 3} + {c \over 3}

Passes through ( $-$ 2, 2), so $-$ 12 = $-$ 8 + c $\Rightarrow$ c = $-$ 4 $\therefore$ 3xy = x3 $-$ 4 i.e. 3x . f(x) = x3 $-$ 4

Q2

The solution of the equation

\,{{{d^2}y} \over {d{x^2}}} = {e^{ - 2x}}

A

{{{e^{ - 2x}}} \over 4}

B

{{{e^{ - 2x}}} \over 4} + cx + d

C

{1 \over 4}{e^{ - 2x}} + c{x^2} + d

D

\,{1 \over 4}{e^{ - 4x}} + cx + d

Correct Answer

Option B

Solution

{{{a^2}y} \over {d{x^2}}} = {e^{ - 2x}};{\mkern 1mu}

{{dy} \over {dx}} = {{{e^{ - 2x}}} \over { - 2}} + c;

y = {{{e^{ - 2x}}} \over 4} + cx + d

Q3

The order and degree of the differential equation

\,{\left( {1 + 3{{dy} \over {dx}}} \right)^{2/3}} = 4{{{d^3}y} \over {d{x^3}}}

are

A

\left( {1,{2 \over 3}} \right)

B

(3, 1)

C

(3,3)

D

(1,2)

Correct Answer

Option C

Solution

{\left( {1 + 3{{d\,y} \over {d\,x}}} \right)^2} =

(4)3

{\left( {{{{d^3}y} \over {d\,{x^3}}}} \right)^3}

\Rightarrow {\left( {1 + 3{{d\,y} \over {d\,x}}} \right)^2} = 16{\left( {{{{d^3}y} \over {d\,{x^3}}}} \right)^3}

$\therefore$ Order = 3 and Degree = 3

Q4

The degree and order of the differential equation of the family of all parabolas whose axis is

x

-axis, are respectively.

A

2, 3

B

2,1

C

1,2

D

3,2.

Correct Answer

Option C

Solution

General equation of parabola whose axis is along X-axis

{y^2} = 4a\left( {x - h} \right)

Differentiating with respect to x, we get

2y{y_1} = 4a

\Rightarrow y{y_1} = 2a

Again differentiating with respect to x, we get

\Rightarrow {y_1}^2 + y{y_2} = 0

Degree

=1,

order

=2.

Q5

The solution of the differential equation

\left( {1 + {y^2}} \right) + \left( {x - {e^{{{\tan }^{ - 1}}y}}} \right){{dy} \over {dx}} = 0,

is :

A

x{e^{2{{\tan }^{ - 1}}y}} = {e^{{{\tan }^{ - 1}}y}} + k

B

\left( {x - 2} \right) = k{e^{2{{\tan }^{ - 1}}y}}

C

2x{e^{{{\tan }^{ - 1}}y}} = {e^{2{{\tan }^{ - 1}}y}} + k

D

x{e^{{{\tan }^{ - 1}}y}} = {\tan ^{ - 1}}y + k

Correct Answer

Option C

Solution

\left( {1 + {y^2}} \right) + \left( {x - {e^{{{\tan }^{ - 1}}y}}} \right){{dy} \over {dx}} = 0

\Rightarrow {{dx} \over {dy}} + {x \over {\left( {1 + {y^2}} \right)}} = {{{e^{{{\tan }^{ - 1}}y}}} \over {\left( {1 + {y^2}} \right)}}

I.F = e{}^{\int {{1 \over {\left( {1 + {y^2}} \right)}}dy} } = {e^{{{\tan }^{ - 1}}y}}

x\left( {{e^{{{\tan }^{ - 1}}y}}} \right) = \int {{{{e^{{{\tan }^{ - 1}}y}}} \over {1 + {y^2}}}} {e^{{{\tan }^{ - 1}}y}}\,dy

x\left( {{e^{{{\tan }^{ - 1}}y}}} \right) = {{{e^{2{{\tan }^{ - 1}}y}}} \over 2} + C

$\therefore$

\,\,\,\,\,2x{e^{{{\tan }^{ - 1}}y}} = {e^{2{{\tan }^{ - 1}}y}} + k

Q6

If a curve

y=f(x)

passes through the point

(1,-1)

and satisfies the differential equation,

y(1+xy) dx=x

dy

, then

f\left( { - {1 \over 2}} \right)

is equal to :

A

{2 \over 5}

B

{4 \over 5}

C

-{2 \over 5}

D

-{4 \over 5}

Correct Answer

Option B

Solution

y\left( {1 + xy} \right)dx = xdy

{{xdy - ydx} \over {{y^2}}} = xdx \Rightarrow \int { - d\left( {{x \over y}} \right) = \int {xdx} }

\,\,\,\,\,\,\,\,\,\,\,\,\,\,

- {x \over y} = {{{x^2}} \over 2} + C\,\,

as

\,\,\,y\left( 1 \right) = - 1 \Rightarrow C = {1 \over 2}

Hence,

y = {{ - 2x} \over {{x^2} + 1}} \Rightarrow f\left( {{{ - 1} \over 2}} \right) = {4 \over 5}

Q7

Solution of the differential equation

ydx + \left( {x + {x^2}y} \right)dy = 0

is

A

log

y=Cx

B

- {1 \over {xy}} + \log y = C

C

{1 \over {xy}} + \log y = C

D

- {1 \over {xy}} = C

Correct Answer

Option B

Solution

ydx + \left( {x + {x^2}y} \right)dy = 0

\Rightarrow {{dx} \over {dy}} = - {x \over y} - {x^2}

\Rightarrow {{dx} \over {dy}} + {x \over y} = - {x^2},

It is Bernoullis form. Divide by

{x^2}

{x^{ - 2}}{{dx} \over {dy}} + {x^{ - 1}}\left( {{1 \over y}} \right) = - 1.

put

{x^{ - 1}} = t,\,\, - {x^{ - 2}}{{dx} \over {dy}} = {{dt} \over {dy}}

We get,

- {{dt} \over {dy}} + t\left( {{1 \over y}} \right) = - 1

\Rightarrow {{dt} \over {dy}} - \left( {{1 \over y}} \right)t = 1

It is linear in

t.

Integrating factor

= {e^{\int { - {1 \over y}dy} }} = {e^{ - \log y}} = {y^{ - 1}}

$\therefore$ Solution is

t\left( {{y^{ - 1}}} \right) = \int {\left( {{y^{ - 1}}} \right)} dy + c

\Rightarrow {1 \over x}.{1 \over y} = \log y + c

\Rightarrow \log y - {1 \over {xy}} = c

Q8

The differential equation for the family of circle

{x^2} + {y^2} - 2ay = 0,

where a is an arbitrary constant is :

A

\left( {{x^2} + {y^2}} \right)y' = 2xy

B

2\left( {{x^2} + {y^2}} \right)y' = xy

C

\left( {{x^2} - {y^2}} \right)y' =2 xy

D

2\left( {{x^2} - {y^2}} \right)y' = xy

Correct Answer

Option C

Solution

{x^2} + {y^2} - 2ay = 0\,\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)

Differentiate,

2x + 2y{{dy} \over {dx}} - 2a{{dy} \over {dx}} = 0

\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Rightarrow a = {{x + yy'} \over {y'}}

Put in

(1),

{x^2} + {y^2} - 2\left( {{{x + yy'} \over {y'}}} \right) = y = 0

\Rightarrow \left( {{x^2} + {y^2}} \right)y' - 2xy - 2{y^2}y' = 0

\Rightarrow \left( {{x^2} - {y^2}} \right)y' = 2xy

Q9

The differential equation representing the family of curves

{y^2} = 2c\left( {x + \sqrt c } \right),

where

c>0,

is a parameter, is of order and degree as follows:

A order

1,

degree

2

B order

1,

degree

1

C order

1,

degree

3

D order

2,

degree

2

Correct Answer

Option C

Solution

{y^2} = 2c\left( {x + \sqrt c } \right)\,\,\,\,\,\,\,\,\,\,...\left( i \right)

2yy' = 2c.1\,\,\,or\,\,\,yy' = c\,\,\,...\left( {ii} \right)

\Rightarrow {y^2} = 2yy'\left( {x + \sqrt {yy'} } \right)

\left[ \, \right.

On putting value of

c

from

(ii)

in

(i)

\left. \, \right]

On simplifying, we get

{\left( {y - 2xy'} \right)^2} = 4yy{'^3}\,\,\,\,\,\,\,\,\,...\left( {iii} \right)

Hence equation

(iii)

is of order

1

and degree

3.

Q10

If

x{{dy} \over {dx}} = y\left( {\log y - \log x + 1} \right),

then the solution of the equation is :

A

y\log \left( {{x \over y}} \right) = cx

B

x\log \left( {{y \over x}} \right) = cy

C

\log \left( {{y \over x}} \right) = cx

D

\log \left( {{x \over y}} \right) = cy

Correct Answer

Option C

Solution

{{xdy} \over {dx}} = y\left( {\log y - \log x + 1} \right)

{{dy} \over {dx}} = {y \over x}\left( {\log \left( {{y \over x}} \right) + 1} \right)

Put

\,\,\,\,y = vx

{{dy} \over {dx}} = v + {{xdv} \over {dx}}

\,\,\,\,\,\,\,\,\,\,\, \Rightarrow v + {{xdv} \over {dx}} = v\left( {\log v + 1} \right)

{{xdv} \over {dx}} = v\,\log \,v

\,\,\,\,\,\,\,\,\,\,\, \Rightarrow {{dv} \over {v\,\log \,v}} = {{dx} \over x}

Put

\,\,\,\,\log \,v = z

\Rightarrow {1 \over v}dv = dz

\Rightarrow {{dz} \over x} = {{dx} \over x}

\Rightarrow \ln \,z = \ln x + \ln \,c

x = cx\,\,\,\,

or

\,\,\,\,\,\log v = cx\,\,\,

or

\,\,\,\,

\log \left( {{y \over x}} \right) = cx.