WB JEE 2008 | Differential Equations Question 1 | Mathematics

WB JEE 2008

MCQ (Single Correct Answer)

-0.25

The function f(x) which satisfies $$f(x) = f( - x) = {{f'(x)} \over x}$$ is given by

$$f(x) = {1 \over 2}{e^{{x^2}}}$$

$$f(x) = {1 \over 2}{e^{ - {x^2}}}$$

$$f(x) = {x^2}{e^{{x^2}/2}}$$

$$f(x) = {e^{{x^2}/2}}$$

WB JEE 2008

MCQ (Single Correct Answer)

-0.25

The degree of the differential equation $${\left[ {1 + {{\left( {{{dy} \over {dx}}} \right)}^2}} \right]^{5/3}} = {{{d^2}y} \over {d{x^2}}}$$ is

10/3

WB JEE 2008

MCQ (Single Correct Answer)

-0.25

The differential equation of all parabolas whose axes are parallel to y-axis is

$${{{d^3}y} \over {d{x^3}}} = 0$$

$${{{d^2}y} \over {d{x^2}}} = 0$$

$${{{d^2}y} \over {d{x^2}}} + {{dy} \over {dx}} = 0$$

$${{{d^2}y} \over {d{x^2}}} + {{dy} \over {dx}} + y = 0$$

WB JEE 2008

MCQ (Single Correct Answer)

-0.25

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

$${e^{ - y}} = {e^x} - {e^{ - x}} + c,\,c$$ integrating constant

$${e^{ - y}} = {e^{ - x}} - {e^x} + c,\,c$$ integrating constant

$${e^{ - y}} = {e^x} + {e^{ - x}} + c,\,c$$ integrating constant

$${e^{ - y}} + {e^x} + {e^{ - x}} = c,\,c$$ integrating constant

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