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1

WB JEE 2008

MCQ (Single Correct Answer)

The differential equation of the family of curves $$y = {e^{2x}}(a\cos x + b\sin x)$$, where a and b are arbitrary constants, is given by

A
y2 $$-$$ 4y1 + 5y = 0
B
2y2 $$-$$ y1 + 5y = 0
C
y2 + 4y1 $$-$$ 5y = 0
D
y2 $$-$$ 2y1 + 5y = 0

Explanation

Given equation of curve is

$$y = {e^{2x}}(a\cos x + b\sin x)$$ ..... (i)

taking derivative w.r.t. x

$${y_1} = {e^{2x}}( - a\sin x + b\cos x) + (a\cos x + b\sin x)2{e^{2x}}$$

or, $${y_1} = {e^{2x}}( - a\sin x + b\cos x) + 2y$$ ..... (ii) (using (i))

Taking derivative again

$${y_2} = 2{e^{2x}}( - a\sin x + b\cos x) + {e^{2x}}( - a\cos x - b\sin x) + 2{y_1}$$

$$ \Rightarrow {y_2} = 2{e^{2x}}( - a\sin x + b\cos x) - y + 2{y_1}$$ (using (i))

$$ \Rightarrow {y_2} = 4{y_1} - 5y$$

or, $${y_2} - 4{y_1} + 5y = 0$$ (using (ii))

2

WB JEE 2008

MCQ (Single Correct Answer)

The differential equation of the family of circles passing through the fixed points (a, 0) and ($$-$$a, 0) is

A
y1(y2 $$-$$ x2) + 2xy + a2 = 0
B
y1y2 + xy + a2x2 = 0
C
y1(y2 $$-$$ x2 + a2) + 2xy = 0
D
y1(y2 + x2) $$-$$ 2xy + a2 = 0

Explanation

Let the equation of circle is

x2 + y2 + 2gx + 2fy + c = 0 ..... (i)

Circle passes through (a, 0) and ($$-$$a, 0)

so a2 + 2ga + c = 0 ..... (ii)

a2 $$-$$ 2ga + c = 0 ...... (iii)

from (ii) and (iii) we get, g = 0 and c = $$-$$a2

$$\therefore$$ x2 + y2 + 2 . 0 . x + 2fy $$-$$ a2 = 0 from (i)

or x2 + y2 + 2fy $$-$$ a2 = 0

or, $$f = {{{a^2} - {x^2} - {y^2}} \over y}$$

Differentiating both sides w.r.t. x, we get

$$0 = {{y( - 2x - 2y'y) - ({a^2} - {x^2} - {y^2})y'} \over {{y^2}}}$$

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

$$ \Rightarrow ({x^2} - {y^2} - {a^2})y' - 2xy = 0$$

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

$$ \Rightarrow {y_1}({y^2} - {x^2} + {a^2}) + 2xy = 0$$ ($$\because$$ $${y_1} = y'$$)

3

WB JEE 2008

MCQ (Single Correct Answer)

The order and degree of the following differential equation $${\left[ {1 + {{\left( {{{dy} \over {dx}}} \right)}^2}} \right]^{5/2}} = {{{d^3}y} \over {d{x^3}}}$$ are respectively

A
3, 2
B
3, 10
C
2, 3
D
3, 5

Explanation

Given differential equation is

$${\left[ {1 + {{\left( {{{dy} \over {dx}}} \right)}^2}} \right]^{5/2}} = {{{d^3}y} \over {d{x^3}}}$$

Squaring both sides

$${\left[ {1 + {{\left( {{{dy} \over {dx}}} \right)}^2}} \right]^5} = {\left( {{{{d^3}y} \over {d{x^3}}}} \right)^2}$$

$$\therefore$$ order = 3, degree = 2.

4

WB JEE 2008

MCQ (Single Correct Answer)

The solution of the differential equation $${{dy} \over {dx}} = {e^{y + x}} + {e^{y - x}}$$ is

A
$${e^{ - y}} = {e^x} - {e^{ - x}} + c,\,c$$ integrating constant
B
$${e^{ - y}} = {e^{ - x}} - {e^x} + c,\,c$$ integrating constant
C
$${e^{ - y}} = {e^x} + {e^{ - x}} + c,\,c$$ integrating constant
D
$${e^{ - y}} + {e^x} + {e^{ - x}} = c,\,c$$ integrating constant

Explanation

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

$${{dy} \over {dx}} = {e^y}({e^x} + {e^{ - x}})$$

$${e^{ - y}}dy = ({e^x} + {e^{ - x}})$$

Integrating both sides we get

$$ \Rightarrow - {e^{ - y}} = {e^x} - {e^{ - x}} + c$$ or $${e^{ - y}} = {e^{ - x}} - {e^x} + c$$.

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