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1

WB JEE 2009

MCQ (Single Correct Answer)

The order of the differential equation $${{{d^2}y} \over {d{x^2}}} = \sqrt {1 + {{\left( {{{dy} \over {dx}}} \right)}^2}} $$ is

A
3
B
2
C
1
D
4

Explanation

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

Order of the differential equation is the highest order derivative in equation. Here $${{{d^2}y} \over {d{x^2}}}$$ is highest order differential coefficient. So order is 2.

2

WB JEE 2009

MCQ (Single Correct Answer)

If x2 + y2 = 1, then

A
yy'' $$-$$ (2y')2 + 1 = 0
B
yy'' + (y')2 + 1 = 0
C
yy'' $$-$$ (y')2 $$-$$ 1 = 0
D
yy'' + 2(y')2 + 1 = 0

Explanation

x2 + y2 = 1

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

2x + 2yy' = 0

$$\Rightarrow$$ x + yy' = 0

Again differentiating both sides, we get

1 + yy'' + y'2 = 0

3

WB JEE 2009

MCQ (Single Correct Answer)

The integrating factor of the differential equation $$x\log x{{dy} \over {dx}} + y = 2\log x$$ is given by

A
ex
B
log x
C
log(log x)
D
x

Explanation

$$x\log x{{dy} \over {dx}} + y = 2\log x$$

$$ \Rightarrow {{dy} \over {dx}} + {y \over {x\log x}} = {2 \over x}$$

$$\therefore$$ Integrating factor $$ = {e^{\int {{1 \over {x\log x}}dx} }} = {e^{\log (\log x)}} = \log x$$

4

WB JEE 2009

MCQ (Single Correct Answer)

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

where c is an arbitrary constant

A
$${e^{ - y}} = {e^x} - {e^{ - x}} + c$$
B
$${e^{ - y}} = {e^{ - x}} - {e^x} + c$$
C
$${e^{ - y}} = {e^x} + {e^{ - x}} + c$$
D
$${e^y} = {e^x} + {e^{ - x}} + c$$

Explanation

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

$$ \Rightarrow {e^{ - y}}dy = ({e^x} + {e^{ - x}})dx$$ (separating variables)

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

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

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

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