1
WB JEE 2020
+1
-0.25
The differential equation of the family of curves y = ex (A cos x + B sin x) where, A, B are arbitrary constants is
A
$${{{d^2}y} \over {d{x^2}}} - 9x = 13$$
B
$${{{d^2}y} \over {d{x^2}}} - 2{{dy} \over {dx}} + 2y = 0$$
C
$${{{d^2}y} \over {d{x^2}}} + 3y = 4$$
D
$${\left( {{{dy} \over {dx}}} \right)^2} + {{dy} \over {dx}} - xy = 0$$
2
WB JEE 2020
+1
-0.25
Let $$y = f(x) = 2{x^2} - 3x + 2$$. The differential of y when x changes from 2 to 1.99 is
A
0.01
B
0.18
C
$$-$$0.05
D
0.07
3
WB JEE 2020
+2
-0.5
Let $$y = {1 \over {1 + x + lnx}}$$, then
A
$$x{{dy} \over {dx}} + y = x$$
B
$$x{{dy} \over {dx}} = y(y\ln x - 1)$$
C
$${x^2}{{dy} \over {dx}} = {y^2} + 1 - {x^2}$$
D
$$x{\left( {{{dy} \over {dx}}} \right)^2} = y - x$$
4
WB JEE 2019
+1
-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)
A
$$x - y{e^{{x \over y}}} = C$$
B
$$y - x{e^{{x \over y}}} = C$$
C
$$x + y{e^{{x \over y}}} = C$$
D
$$y + x{e^{{x \over y}}} = C$$
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