1
WB JEE 2024
+1
-0.25

If $$x y^{\prime}+y-e^x=0, y(a)=b$$, then $$\lim _\limits{x \rightarrow 1} y(x)$$ is

A
$$e+2 a b-e^a$$
B
$$e^2+a b-e^{-a}$$
C
$$\mathrm{e}-\mathrm{ab}+\mathrm{e}^{\mathrm{a}}$$
D
$$\mathrm{e}+\mathrm{ab}-\mathrm{e}^{\mathrm{a}},\left(\mathrm{y}^{\prime}=\frac{\mathrm{dy}}{\mathrm{d} x}\right)$$
2
WB JEE 2023
+1
-0.25

If $$y = {x \over {{{\log }_e}|cx|}}$$ is the solution of the differential equation $${{dy} \over {dx}} = {y \over x} + \phi \left( {{x \over y}} \right)$$, then $$\phi \left( {{x \over y}} \right)$$ is given by

A
$${{{y^2}} \over {{x^2}}}$$
B
$$- {{{y^2}} \over {{x^2}}}$$
C
$${{{x^2}} \over {{y^2}}}$$
D
$$- {{{x^2}} \over {{y^2}}}$$
3
WB JEE 2023
+1
-0.25

Given $${{{d^2}y} \over {d{x^2}}} + \cot x{{dy} \over {dx}} + 4y\cos e{c^2}x = 0$$. Changing the independent variable x to z by the substitution $$z = \log \tan {x \over 2}$$, the equation is changed to

A
$${{{d^2}y} \over {d{z^2}}} + {3 \over y} = 0$$
B
$$2{{{d^2}y} \over {d{z^2}}} + {e^y} = 0$$
C
$${{{d^2}y} \over {d{z^2}}} - 4y = 0$$
D
$${{{d^2}y} \over {d{z^2}}} + 4y = 0$$
4
WB JEE 2023
+2
-0.5

The family of curves $$y = {e^{a\sin x}}$$, where 'a' is arbitrary constant, is represented by the differential equation

A
$$y\log y = \tan x{{dy} \over {dx}}$$
B
$$y\log x = \cot x{{dy} \over {dx}}$$
C
$$\log y = \tan x{{dy} \over {dx}}$$
D
$$\log y = \cot x{{dy} \over {dx}}$$
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