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1

WB JEE 2009

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

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

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

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)

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

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

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

On those following papers in MCQ (Single Correct Answer)
Number in Brackets after Paper Indicates No. of Questions
WB JEE 2022 (2)
WB JEE 2021 (3)
WB JEE 2020 (6)
WB JEE 2019 (2)
WB JEE 2018 (2)

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