WB JEE 2009 | Differential Equations Question 11 | Mathematics

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 2009

MCQ (Single Correct Answer)

-0.25

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

WB JEE 2010

MCQ (Single Correct Answer)

-0.25

The general solution of the differential equation $$100{{{d^2}y} \over {d{x^2}}} - 20{{dy} \over {dx}} + y = 0$$ is

$$y = ({c_1} + {c_2}x){e^x}$$

$$y = ({c_1} + {c_2}x){e^{ - x}}$$

$$y = ({c_1} + {c_2}x){e^{{x \over {10}}}}$$

$$y = {c_1}{e^x} + {c_2}{e^{ - x}}$$

WB JEE 2010

MCQ (Single Correct Answer)

-0.25

If $$y'' - 3y' + 2y = 0$$ where y(0) = 1, y'(0) = 0, then the value of y at $$x = {\log _e}2$$ is

$$-$$1

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