WB JEE 2020 | Differential Equations Question 48 | Mathematics

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WB JEE 2020

MCQ (Single Correct Answer)

-0.25

Let $$y = f(x) = 2{x^2} - 3x + 2$$. The differential of y when x changes from 2 to 1.99 is

0.01

0.18

$$ - $$0.05

0.07

WB JEE 2020

MCQ (Single Correct Answer)

-0.5

Let $$y = {1 \over {1 + x + lnx}}$$, then

$$x{{dy} \over {dx}} + y = x$$

$$x{{dy} \over {dx}} = y(y\ln x - 1)$$

$${x^2}{{dy} \over {dx}} = {y^2} + 1 - {x^2}$$

$$x{\left( {{{dy} \over {dx}}} \right)^2} = y - x$$

WB JEE 2019

MCQ (Single Correct Answer)

-0.25

The general solution of the differential equation $$\left( {1 + {e^{{x \over y}}}} \right)dx + \left( {1 - {x \over y}} \right){e^{x/y}}dy = 0$$ is (C is an arbitrary constant)

$$x - y{e^{{x \over y}}} = C$$

$$y - x{e^{{x \over y}}} = C$$

$$x + y{e^{{x \over y}}} = C$$

$$y + x{e^{{x \over y}}} = C$$

WB JEE 2019

MCQ (Single Correct Answer)

-0.25

General solution of $${(x + y)^2}{{dy} \over {dx}} = {a^2},a \ne 0$$ is (C is an arbitrary constant)

$${x \over a} = \tan {y \over a} + C$$

$$\tan xy = C$$

$$\tan (x + y) = C$$

$$\tan {{y + C} \over a} = {{x + y} \over a}$$

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