Differential Equations — Page 9 | JEE Mathematics

Q81

Let the slope of the tangent to a curve y = f(x) at (x, y) be given by 2

\tan x(\cos x - y)

. If the curve passes through the point

\left( {{\pi \over 4},0} \right)

, then the value of

\int\limits_0^{\pi /2} {y\,dx}

is equal to :

A

(2 - \sqrt 2 ) + {\pi \over {\sqrt 2 }}

B

2 - {\pi \over {\sqrt 2 }}

C

(2 + \sqrt 2 ) + {\pi \over {\sqrt 2 }}

D

2 + {\pi \over {\sqrt 2 }}

Correct Answer

Option B

Solution

{{dy} \over {dx}} = 2\tan x(\cos x - y)

\Rightarrow {{dy} \over {dx}} + 2\tan xy = 2\sin x

I.F. = {e^{\int {2\tan xdx} }} = {\sec ^2}x

$\therefore$ Solution of D.E. will be

y(x){\sec ^2}x = \int {2\sin x{{\sec }^2}xdx}

y{\sec ^2}x = 2\sec x + c

$\because$ Curve passes through

\left( {{\pi \over 4},0} \right)

$\therefore$

c = - 2\sqrt 2

$\therefore$

y = 2\cos x - 2\sqrt 2 {\cos ^2}x

$\therefore$

\int_0^{\pi /2} {ydx = \int_0^{\pi /2} {(2\cos x - 2\sqrt 2 {{\cos }^2}x)\,dx} }

= 2 - 2\sqrt 2 \,.\,{\pi \over 4} = 2 - {\pi \over {\sqrt 2 }}

Q82

If

y = y(x)

is the solution of the differential equation

2{x^2}{{dy} \over {dx}} - 2xy + 3{y^2} = 0

such that

y(e) = {e \over 3}

, then y(1) is equal to :

A

{1 \over 3}

B

{2 \over 3}

C

{3 \over 2}

D 3

Correct Answer

Option B

Solution

2{x^2}{{dy} \over {dx}} - 2xy + 3{y^2} = 0

\Rightarrow 2x(xdy - ydx) + 3{y^3}dx = 0

\Rightarrow 2\left( {{{xdy - ydx} \over {{y^2}}}} \right) + 3{{dx} \over x} = 0

\Rightarrow - {{2x} \over y} + 3\ln x = C

$\because$

y(e) = {e \over 3} \Rightarrow - 6 + 3 = C \Rightarrow C = - 3

Now, at

x = 1

,

- {2 \over y} + 0 = - 3

y = {2 \over 3}

Q83

Let the solution curve

y = y(x)

of the differential equation

\left[ {{x \over {\sqrt {{x^2} - {y^2}} }} + {e^{{y \over x}}}} \right]x{{dy} \over {dx}} = x + \left[ {{x \over {\sqrt {{x^2} - {y^2}} }} + {e^{{y \over x}}}} \right]y

pass through the points (1, 0) and (2

\alpha

,

\alpha

),

\alpha

> 0. Then

\alpha

is equal to

A

{1 \over 2}\exp \left( {{\pi \over 6} + \sqrt e - 1} \right)

B

{1 \over 2}\exp \left( {{\pi \over 6} + e - 1} \right)

C

\exp \left( {{\pi \over 6} + \sqrt e + 1} \right)

D

2\exp \left( {{\pi \over 3} + \sqrt e - 1} \right)

Correct Answer

Option A

Solution

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

Putting y = tx

\left( {{1 \over {\sqrt {1 - {t^2}} }} + {e^t}} \right)\left( {t + x{{dt} \over {dx}}} \right) = 1 + \left( {{1 \over {\sqrt {1 - {t^2}} }} + {e^t}} \right)t

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

\Rightarrow {\sin ^{ - 1}}t + {e^t} = \ln x + C

\Rightarrow {\sin ^{ - 1}}\left( {{y \over x}} \right) + {e^{y/x}} = \ln x + C

at x = 1, y = 0 So,

0 + {e^0} = 0 + C \Rightarrow C = 1

at

(2\alpha ,\alpha )

{\sin ^{ - 1}}\left( {{y \over x}} \right) + {e^{y/x}} = \ln x + 1

\Rightarrow {\pi \over 6} + {e^{{1 \over 2}}} - 1 = \ln (2\alpha )

\Rightarrow \alpha = {1 \over 2}{e^{\left( {{\pi \over 6} + {e^{{1 \over 2}}} - 1} \right)}}

Q84

If the solution curve of the differential equation

(({\tan ^{ - 1}}y) - x)dy = (1 + {y^2})dx

passes through the point (1, 0), then the abscissa of the point on the curve whose ordinate is tan(1), is

A 2e

B

{2 \over e}

C 2

D

{1 \over e}

Correct Answer

Option B

Solution

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

{{dx} \over {dy}} + {x \over {1 + {y^2}}} = {{{{\tan }^{ - 1}}y} \over {1 + {y^2}}}

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

$\therefore$ Solution

x.\,{e^{{{\tan }^{ - 1}}y}} = \int {{{{e^{{{\tan }^{ - 1}}y}}{{\tan }^{ - 1}}y} \over {1 + {y^2}}}dy}

Let

{e^{{{\tan }^{ - 1}}y}} = t

{{{e^{{{\tan }^{ - 1}}y}}} \over {1 + {y^2}}} = dt

= x{e^{{{\tan }^{ - 1}}y}} = \int {\ln tdt = t\ln t - t + c}

$\therefore$

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

..... (i) $\because$ It passes through (1, 0) $\Rightarrow$ c = 2 Now put y = tan1, then

ex = e - e + 2

\Rightarrow x = {2 \over e}

Q85

Let

{{dy} \over {dx}} = {{ax - by + a} \over {bx + cy + a}}

, where a, b, c are constants, represent a circle passing through the point (2, 5). Then the shortest distance of the point (11, 6) from this circle is :

A 10

B 8

C 7

D 5

Correct Answer

Option B

Solution

{{dy} \over {dx}} = {{ax - by + a} \over {bx + cy + a}}

= bx\,dy + cy\,dy + a\,dy = ax\,dx - by\,dx + a\,dx

= cy\,dy + a\,dy - ax\,dx - a\,dx + b(x\,dy + y\,dx) = 0

= c\int {y\,dy + a\int {x\,dx - a\int {dx + b\int {d(xy) = 0} } } }

= {{c{y^2}} \over 2} + ay - {{a{x^2}} \over 2} - ax + bxy = k

= a{x^2} - c{y^2} + 2ax - 2ay - 2bxy = k

Above equation is circle $\Rightarrow$ a = $-$ c and b = 0

a{x^2} + a{y^2} + 2ax - 2ay = k

\Rightarrow {x^2} + {y^2} + 2x - 2y = \lambda \,\,\,\,\,\,\,\left[ {\lambda = {k \over a}} \right]

Passes through (2, 5)

4 + 25 + 4 - 10 = \lambda \Rightarrow \lambda = 23

Circle

\equiv {x^2} + {y^2} + 2x - 2y - 23 = 0

Centre ( $-$ 1, 1)

r = \sqrt {{{( - 1)}^2} + {1^2} + 23} = 5

Shortest distance of

(11,6) = \sqrt {{{12}^2} + {5^2}} - 5

= 13 - 5

= 8

Q86

If

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

, x, y > 0, y(1) = 1, then y(2) is equal to :

A

2 + {\log _2}3

B

2 + {\log _3}2

C

2 - {\log _3}2

D

2 - {\log _2}3

Correct Answer

Option D

Solution

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

, x, y > 0, y(1) = 1

{{dy} \over {dx}} = - {{{2^x}({2^y} - 1)} \over {{2^y}({2^x} - 1)}}

\int {{{{2^y}} \over {{2^y} - 1}}dy = - \int {{{{2^x}} \over {{2^x} - 1}}dx} }

= {{{{\log }_e}({2^y} - 1)} \over {{{\log }_e}2}} = - {{{{\log }_e}({2^x} - 1)} \over {{{\log }_e}2}} + {{{{\log }_e}c} \over {{{\log }_e}2}}

= |({2^y} - 1)({2^x} - 1)| = c

$\because$

y(1) = 1

$\therefore$

c = 1

= |({2^y} - 1)({2^x} - 1)| = 1

For

x = 2

|({2^y} - 1)3| = 1

{2^y} - 1 = {1 \over 3} \Rightarrow 2y = {4 \over 3}

Taking log to base 2. $\therefore$

y = 2 - {\log _2}3

Q87

If

y = y(x)

is the solution of the differential equation

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

,

y(1) = 0

then the local maximum value of the function

z(x) = {x^2}y(x) - {e^x},\,x \in R

is :

A 1

-

e

B 0

C

{1 \over 2}

D

{4 \over e} - e

Correct Answer

Option D

Solution

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

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

, then

{e^{\int {{2 \over x}dx} }}dx = {x^2}

y\,.\,{x^2} = \int {{x^2}{e^x}dx}

y{x^2} = {x^2}{e^x} - \int {2x{e^x}dx}

= {x^2}{e^x} - 2(x{e^x} - {e^x}) + c

y{x^2} = {x^2}{e^x} - 2x{e^x} + 2{e^x} + c

y{x^2} = ({x^2} - 2x + 2){e^x} + c

0 = e + c \Rightarrow c = - e

y(x)\,.\,{x^2} - {e^x} = {(x - 1)^2}{e^x} - e

z(x) = {(x - 1)^2}{e^x} - e

For local maximum

z'(x) = 0

$\therefore$

2(x - 1){e^x} + {(x - 1)^2}{e^x} = 0

$\therefore$

x = - 1

And local maximum value

= z( - 1)

= {4 \over e} - e

Q88

Let

y = y(x)

be the solution of the differential equation

{x^3}dy + (xy - 1)dx = 0,x > 0,y\left( {{1 \over 2}} \right) = 3 - \mathrm{e}

. Then y (1) is equal to

A 2

-

e

B 3

C 1

D e

Correct Answer

Option C

Solution

$x^{3} d y+x y d x-d x=0$ $\Rightarrow \dfrac{d y}{d x}=\dfrac{1-x y}{x^{3}}$ $\Rightarrow \dfrac{d y}{d x}+\dfrac{y}{x^{2}}=\dfrac{1}{x^{3}}$ I.F.

$=e^{\int \dfrac{d x}{x^{2}}}=e^{-\dfrac{1}{x}}$ $\therefore \quad y e^{-\dfrac{1}{x}}=\int \dfrac{e^{-\dfrac{1}{x}}}{x^{3}} d x$ For RHS put $-\dfrac{1}{x}=t \Rightarrow \dfrac{d x}{x^{2}}=d t$ $\therefore y e^{-\dfrac{1}{x}}=-\int t e^{t} d t$ $\Rightarrow y e^{-\dfrac{1}{x}}=-\left[t e^{t}-e^{t}\right]+c$ $\Rightarrow y e^{-\dfrac{1}{x}}=\dfrac{e^{-\dfrac{1}{x}}}{x}+e^{-\dfrac{1}{x}}+c$

\downarrow y\left(\frac{1}{2}\right)=3-e

$\Rightarrow(3-e) e^{-2}=2 e^{-2}+e^{-2}+c$ $\Rightarrow \quad c=-\dfrac{1}{e}$ For $y(1)$ put $x=1, c=-e^{-1}$ in equation (i) we get $y e^{-1}=e^{-1}+e^{-1}-e^{-1}$ $\Rightarrow y=1$

Q89

Let

g:(0,\infty ) \to R

be a differentiable function such that

\int {\left( {{{x(\cos x - \sin x)} \over {{e^x} + 1}} + {{g(x)\left( {{e^x} + 1 - x{e^x}} \right)} \over {{{({e^x} + 1)}^2}}}} \right)dx = {{x\,g(x)} \over {{e^x} + 1}} + c}

, for all x > 0, where c is an arbitrary constant. Then :

A g is decreasing in

\left( {0,{\pi \over 4}} \right)

B g' is increasing in

\left( {0,{\pi \over 4}} \right)

C g + g' is increasing in

\left( {0,{\pi \over 2}} \right)

D g

-

g' is increasing in

\left( {0,{\pi \over 2}} \right)

Correct Answer

Option D

Solution

\int\left(\frac{x(\cos x-\sin x)}{e^x+1}+\frac{g(x)\left(e^x+1-x e^x\right)}{\left(e^x+1\right)^2}\right) d x=\frac{x g(x)}{e^x+1}+c

On differentiating both sides w.r.t. $\mathrm{x}$ , we get

\begin{aligned} &\left(\frac{x(\cos x-\sin x)}{e^x+1}+\frac{g(x)\left(e^x+1-x e^x\right.}{\left(e^x+1\right)^2}\right) \\\\ &=\frac{\left(e^x+1\right)\left(g(x)+x g^{\prime}(x)\right)-e^x \cdot x \cdot g(x)}{\left(e^x+1\right)^2} \\\\ &\left(e^x+1\right) x(\cos x-\sin x)+g(x)\left(e^x+1-x e^x\right) \\\\ &=\left(e^x+1\right)\left(g(x)+x g^{\prime}(x)\right)-e^x \cdot x \cdot g(x) \\\\ &\Rightarrow g g^{\prime}(x)=\cos x-\sin x \\\\ &\Rightarrow g(x)=\sin x+\cos x+C \end{aligned}

$\mathrm{g}(\mathrm{x})$ is increasing in $(0, \pi / 4)$ $g "(x)=-\sin x-\cos x0$ \Rightarrow \phi$ is increasing Hence option D is correct.

Q90

Let a smooth curve

y=f(x)

be such that the slope of the tangent at any point

(x, y)

on it is directly proportional to

\left(\dfrac{-y}{x}\right)

. If the curve passes through the points

(1,2)

and

(8,1)

, then

\left|y\left(\dfrac{1}{8}\right)\right|

is equal to

A

2 \log _{e} 2

B 4

C 1

D

4 \log _{e} 2

Correct Answer

Option B

Solution

{{dy} \over {dx}} \propto {{ - y} \over x}

{{dy} \over {dx}} = {{ - ky} \over x} \Rightarrow \int {{{dy} \over y} = - K\int {{{dx} \over x}} }

\ln |y| = - K\ln |x| + C

If the above equation satisfy (1, 2) and (8, 1)

\ln 2 = - K \times 0 + C \Rightarrow C = \ln 2

\ln 1 = - K\ln 8 + \ln 2 \Rightarrow K = {1 \over 3}

So, at

x = {1 \over 8}

\ln |y| = - {1 \over 3}\ln \left( {{1 \over 8}} \right) + \ln 2 = 2\ln 2

|y| = 4