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1

### 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

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

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

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

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

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

2

### 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.

3

### 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 $${e^{ - y}} = {e^{ - x}} - {e^x} + c$$.

4

### WB JEE 2008

MCQ (Single Correct Answer)

The differential equation of all parabolas whose axes are parallel to y-axis is

A
$${{{d^3}y} \over {d{x^3}}} = 0$$
B
$${{{d^2}y} \over {d{x^2}}} = 0$$
C
$${{{d^2}y} \over {d{x^2}}} + {{dy} \over {dx}} = 0$$
D
$${{{d^2}y} \over {d{x^2}}} + {{dy} \over {dx}} + y = 0$$

## Explanation

Let equation of parabola whose axis parallel to y-axis, is $${(x - \alpha )^2} = 4a(y - \beta )$$ where ($$\alpha$$, $$\beta$$) is vertex of parabola

differentiating w.r.t. x

$$2(x - \alpha ) = 4a{{dy} \over {dx}}$$ or $$(x - \alpha ) = 2a{{dy} \over {dx}}$$

Taking derivative again

$$1 = 2a{{{d^2}y} \over {d{x^2}}}$$ or $${{{d^2}y} \over {d{x^2}}} = {1 \over {2a}}$$

Taking derivative again

$${{{d^3}y} \over {d{x^3}}} = 0$$.

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