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1

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

2

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)

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

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

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

3

WB JEE 2009

MCQ (Single Correct Answer)

The slope at any point of a curve y = f(x) is given by $${{dy} \over {dx}} = 3{x^2}$$ and it passes through ($$-$$1, 1). The equation of the curve is

A
y = x3 + 2
B
y = $$-$$x3 + 4
C
y = 3x2 + 4
D
y = $$-$$x3 $$-$$ 2

Explanation

$$\int {dy = \int {3{x^2}dx \Rightarrow y = {x^3} + c} } $$

$$\because$$ The curve passes through ($$-$$1, 1)

$$1 = - 1 + c \Rightarrow c = 2$$

$$y = {x^3} + 2$$.

4

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))

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