WB JEE 2009 | Differential Equations Question 9 | Mathematics

WB JEE 2009

MCQ (Single Correct Answer)

-0.25

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

y = x³ + 2

y = $$-$$x³ + 4

y = 3x² + 4

y = $$-$$x³ $$-$$ 2

WB JEE 2009

MCQ (Single Correct Answer)

-0.25

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

where c is an arbitrary constant

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

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

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

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

WB JEE 2009

MCQ (Single Correct Answer)

-0.25

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

e^x

log x

log(log x)

WB JEE 2009

MCQ (Single Correct Answer)

-0.25

If x² + y² = 1, then

yy'' $$-$$ (2y')² + 1 = 0

yy'' + (y')² + 1 = 0

yy'' $$-$$ (y')² $$-$$ 1 = 0

yy'' + 2(y')² + 1 = 0

WB JEE Subjects