Differential Equations — Page 18 | JEE Mathematics

Q171

Let y = y(x) be the solution of the differential equation, x

{{dy} \over {dx}}

+ y = x loge x, (x > 1). If 2y(2) = loge 4

-

1, then y(e) is equal to :

A

- {e \over 2}

B

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

C

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

D

{e \over 4}

Correct Answer

Option D

Solution

{{dy} \over {dx}} = {y \over x} = \ell nx

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

xy = \int {x\ell nx + C}

\ell nx{{{x^2}} \over 2} - \int {{1 \over x}.{{{x^2}} \over 2}}

xy = {x \over 2}\ell nx - {{{x^2}} \over 4} + C,

for

2y\left( 2 \right) = 2\ell n2 - 1

$\Rightarrow$

C = 0

y = {x \over 2}\ell nx - {x \over 4}

y\left( e \right) = {e \over 4}

Q172

The solution of the differential equation,

{{dy} \over {dx}}

= (x – y)2, when y(1) = 1, is :

A

-

loge

\left| {{{1 + x - y} \over {1 - x + y}}} \right|

= x + y

-

2

B loge

\left| {{{2 - x} \over {2 - y}}} \right|

= x

-

y

C loge

\left| {{{2 - y} \over {2 - x}}} \right|

= 2(y

-

1)

D

-

loge

\left| {{{1 - x + y} \over {1 + x - y}}} \right|

= 2(x

-

1)

Correct Answer

Option D

Solution

x $-$ y = t $\Rightarrow$

{{dy} \over {dx}} = 1 - {{dt} \over {dx}}

$\Rightarrow$ 1 $-$

{{dt} \over {dx}}

= t2 $\Rightarrow$

\int {{{dt} \over {1 - {t^2}}}}

=

\int {1dx}

$\Rightarrow$

{1 \over 2}\ell n\left( {{{1 + t} \over {1 - t}}} \right) = x + \lambda

$\Rightarrow$

{1 \over 2}\ell n\left( {{{1 + x - y} \over {1 - x + y}}} \right) = x + \lambda

given y(1) = 1 $\Rightarrow$

{1 \over 2}\ell n(1) = 1 + \lambda \Rightarrow \lambda = - 1

$\Rightarrow$

\ell n\left( {{{1 + x - y} \over {1 - x + y}}} \right)

= 2(x $-$ 1) $\Rightarrow$

- \ell n\left( {{{1 - x + y} \over {1 + x - y}}} \right)

= 2(x $-$ 1)

Q173

Let f : [0,1]

\to

R be such that f(xy) = f(x).f(y), for all x, y

\in

[0, 1], and f(0)

\ne

0. If y = y(x) satiesfies the differential equation,

{{dy} \over {dx}}

= f(x) with y(0) = 1, then y

\left( {{1 \over 4}} \right)

+ y

\left( {{3 \over 4}} \right)

is equal to :

A 3

B 4

C 2

D 5

Correct Answer

Option A

Solution

If f(xy) = f(x) f(y) $\forall$ x, y

\in

R and f(0) $\ne$ 0 put x = y = 0 $\Rightarrow$ f(0) = [f(0)]2 $\Rightarrow$ f(0) = 1 put y = 0 $\Rightarrow$ f(0) = f(x) f(0) $\Rightarrow$ f(x) = 1 given that

{{dy} \over {dx}}

= f(x) $\therefore$

{{dy} \over {dx}}

= 1 $\Rightarrow$ y = x + k given that y(0) = 1 $\therefore$ k = 1 hence y = x + 1 y

\left( {{1 \over 4}} \right)

+ y

\left( {{3 \over 4}} \right)

=

\left( {{1 \over 4} + 1} \right)

+

\left( {{3 \over 4} + 1} \right)

= 3

Q174

Let f be a differentiable function such that f '(x) = 7 -

{3 \over 4}{{f\left( x \right)} \over x},

(x > 0) and f(1)

\ne

4. Then

\mathop {\lim }\limits_{x \to 0'} \,

xf

\left( {{1 \over x}} \right)

:

A does not exist

B exists and equals

{4 \over 7}

C exists and equals 4

D exists and equals 0

Correct Answer

Option C

Solution

f'(x) = 7 - {3 \over 4}{{f\left( x \right)} \over x}\,\,\,\left( {x > 0} \right)

Given f(1) $\ne$ 4

\mathop {\lim }\limits_{x \to {0^ + }} \,xf\left( {{1 \over x}} \right)\, = ?

{{dy} \over {dx}} + {3 \over 4}{y \over x} = 7

(This is LDE) IF

= {e^{\int {{3 \over {4x}}dx} }} = {e^{{3 \over 4}\ln \left| x \right|}} = {x^{{3 \over 4}}}

y.{x^{{3 \over 4}}} = \int {7.{x^{{3 \over 4}}}} dx

y.{x^{{3 \over 4}}} = 7.{{{x^{{7 \over 4}}}} \over {{7 \over 4}}} + C

f(x) = 4x + C.{x^{ - {3 \over 4}}}

f\left( {{1 \over 4}} \right) = {4 \over x} + C.{x^{{3 \over 4}}}

\mathop {\lim }\limits_{x \to {0^ + }} xf\left( {{1 \over x}} \right) = \mathop {\lim }\limits_{x \to {0^ + }} \left( {4 + C.{x^{{7 \over 4}}}} \right) = 4

Q175

The curve amongst the family of curves represented by the differential equation, (x2 – y2)dx + 2xy dy = 0 which passes through (1, 1) is :

A a circle with centre on the y-axis

B an ellipse with major axis along the y-axis

C a circle with centre on the x-axis

D a hyperbola with transverse axis along the x-axis

Correct Answer

Option C

Solution

(x2 $-$ y2) dx + 2xy dy = 0

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

Put

y = vx \Rightarrow {{dy} \over {dx}} = v + x{{dv} \over {dx}}

Solving we get,

\int {{{2v} \over {{v^2} + 1}}dv = \int { - {{dx} \over x}} }

ln(v2 + 1) = $-$ ln x + C (y2 + x2) = Cx 1 + 1 = C $\Rightarrow$ C = 2 y2 + x2 = 2x

Q176

The solution of the differential equation

{{dy} \over {dx}}\, + \,{y \over 2}\,\sec x = {{\tan x} \over {2y}},\,\,

where 0

\le

x <

{\pi \over 2}

, and y (0) = 1, is given by :

A y = 1

-

{x \over {\sec x + \tan x}}

B y2 = 1 +

{x \over {\sec x + \tan x}}

C y2 = 1

-

{x \over {\sec x + \tan x}}

D y = 1 +

{x \over {\sec x + \tan x}}

Correct Answer

Option C

Solution

Given,

{{dy} \over {dx}} + {y \over 2}\sec x = {{\tan x} \over {2y}}

$\Rightarrow$

2y{{dy} \over {dx}} + {y^2}\sec x = \tan x

Now, let y2 $=$ t $\Rightarrow$ 2y

{{dy} \over {dx}} = {{dt} \over {dx}}

$\therefore$ New equation,

{{dt} \over {dx}} + t\sec x = \tan x

$\therefore$ I.F $=$

{e^{\int {\sec xdx} }}

$=$

{e^{\ln \left( {\sec x + \tan x} \right)}}

$=$ sec x + tan x $\therefore$ Solution is, t(sec x + tan x) $=$

\int {\tan x}

(sec x + tan x) dx $\Rightarrow$ t(sec x + tan x) $=$ sec x + tan x $-$ x + c $\Rightarrow$ t $=$ 1 $-$

{x \over {\sec x + \tan x}} + c

$\Rightarrow$ y2 $=$ 1 $-$

{x \over {\sec x + \tan x}} + c

Given, y(0) $=$ 1 $\therefore$ 1 $=$ 1 $-$ 0 + c $\Rightarrow$ c $=$ 0 $\therefore$ y2 $=$ 1 $-$

{x \over {\sec x + \tan x}}

Q177

The curve satisfying the differential equation, ydx

-

(x + 3y2)dy = 0 and passing through the point (1, 1), also passes through the point :

A

\left( {{1 \over 4}, - {1 \over 2}} \right)

B

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

C

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

D

\left( {{1 \over 4}, {1 \over 2}} \right)

Correct Answer

Option B

Solution

Given, y dx =

\left( {x + 3{y^2}} \right)dy

$\Rightarrow$

\,\,\,

y

{{dx} \over {dy}}

= x + 3y2 $\Rightarrow$

\,\,\,

{{dx} \over {dy}}

$-$

{x \over y} = 3y

If =

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

=

{e^{ - \ln y}}

=

{1 \over y}

\therefore\,\,\,

Soluation is , x .

{1 \over y}

=

\int {3y.{1 \over y}dy}

$\Rightarrow$

\,\,\,

{x \over y}

= 3y + c This curve passing through (1, 1)

\therefore\,\,\,

1 = 3 + c $\Rightarrow$

\,\,\,

c = $-$ 2

\therefore\,\,\,

Curve is, x = 3y2 $-$ 2y Now put every point in this equation, and see which point satisfy this equation.

Following this method you can see ( $-$

{1 \over 3}

,

{1 \over 3}

) point satisfy this equation.

Q178

If f(x) is a differentiable function in the interval (0,

\infty

) such that f (1) = 1 and

\mathop {\lim }\limits_{t \to x}

{{{t^2}f\left( x \right) - {x^2}f\left( t \right)} \over {t - x}} = 1,

for each x > 0, then

f\left( {{{3}/{2}}} \right)

equal to :

A

{{13} \over 6}

B

{{23} \over 18}

C

{{25} \over 9}

D

{{31} \over 18}

Correct Answer

Option D

Solution

\mathop {\lim }\limits_{t \to x} {{{t^2}f\left( x \right) - {x^2}f\left( t \right)} \over {t - x}} = 1

It is in

{0 \over 0}

form So, applying L' Hospital rule,

\mathop {\lim }\limits_{t \to x} {{2tf\left( x \right) - {x^2}f'\left( x \right)} \over 1} = 1

$\Rightarrow$ 2xf(x) $-$ x2f '(x) = 1 $\Rightarrow$ f '(x) $-$

{2 \over x}

f(x) $=$

{1 \over {{x^2}}}

$\therefore$ I.F =

{e^{\int {{{ - 2} \over x}dx} }} = {e^{ - 2\log x}} = {1 \over {{x^2}}}

$\therefore$ Solution of equation, f(x)

{1 \over {{x^2}}}

=

\int {{1 \over x}\left( { - {1 \over {{x^2}}}} \right)} \,dx

$\Rightarrow$

{{f(x)} \over {{x^2}}} = {1 \over {3{x^3}}} + C

Given that, f(1) = 1 $\therefore$

{1 \over 1}

=

{1 \over 3}

+ C $\Rightarrow$ C $=$

{2 \over 3}

$\therefore$ f(x) $=$

{2 \over 3}

x2 +

{1 \over {3x}}

$\therefore$ f

\left( {{3 \over 2}} \right) = {2 \over 3} \times {\left( {{3 \over 2}} \right)^2} + {1 \over 3} \times {2 \over 3} = {{31} \over {18}}

Q179

If y(x) is the solution of the differential equation

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

where

y\left( 1 \right) = {1 \over 2}{e^{ - 2}},

then

A y(loge2) = loge4

B y(x) is decreasing in (0, 1)

C y(loge2) =

{{{{\log }_e}2} \over 4}

D y(x) is decreasing in

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

Correct Answer

Option D

Solution

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

I.F.

= {e^{\int {\left( {{{2x + 1} \over x}} \right)dx} }} = {e^{\int {\left( {2 + {1 \over x}} \right)dx} }} = {e^{2x + \ell nx}} = {e^{2x}}.x

So,

y\left( {x{e^{2x}}} \right) = \int {{e^{ - 2x}}.x{e^{2x}} + C}

\Rightarrow xy{e^{2x}} = \int {xdx + C}

\Rightarrow 2xy{e^{2x}} = {x^2} + 2C

It passes through

\left( {1,{1 \over 2}{e^{ - 2}}} \right)

we get C $=$ 0

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

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

\Rightarrow f(x)

is decreasing in

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

y\left( {{{\log }_e}2} \right) = {{\left( {{{\log }_e}2} \right){e^{ - 2({{\log }_e}2)}}} \over 2}

= {1 \over 8}{\log _e}2