1
BITSAT 2024
MCQ (Single Correct Answer)
+3
-1

The solution of the differential equation

$ (x+1) \frac{d y}{d x}-y=e^{3 x}(x+1)^{2} $ is

A
$ y=(x+1) e^{3 x}+C $
B
$ 3 y=(x+1)+e^{3 x}+C $
C
$ \frac{3 y}{x+1}=e^{3 x}+C $
D
$ y e^{-3 x}=3(x+1)+C $
2
BITSAT 2023
MCQ (Single Correct Answer)
+3
-1

If $$\left(1+x^2\right) d y+2 x y d x=\cot x d x$$, then the general solution be

A
$$y=\frac{\log |\sin x|}{1+x^2}+\frac{C}{1+x^2}$$
B
$$y=\frac{\log |\sin x|}{1-x^2}+\frac{C}{1-x^2}$$
C
$$y=\frac{\log |\cos x|}{1+x^2}+\frac{C}{1+x^2}$$
D
$$y=\frac{\log |\cos x|}{1-x^2}+\frac{C}{1-x^2}$$
3
BITSAT 2022
MCQ (Single Correct Answer)
+3
-1

$$\left( {{{dy} \over {dx}}} \right)\tan x = y{\sec ^2}x + \sin x$$, find general solution

A
$$y = \tan x(\log |{\mathop{\rm cosec}\nolimits} x - \cot x| + \cos x + c)$$
B
$$y = {\sec ^2}x + \tan x + c$$
C
$$y = \log |\sec x + \tan x| + {\mathop{\rm cosec}\nolimits} \,x + c$$
D
$$y = {\tan ^2}x + \sin x + c$$
4
BITSAT 2021
MCQ (Single Correct Answer)
+3
-1

Solution of $$\left( {{{x + y - 1} \over {x + y - 2}}} \right){{dy} \over {dx}} = \left( {{{x + y + 1} \over {x + y + 2}}} \right)$$, given that y = 1 when x = 1 is

A
$$\ln \left| {{{{{(x - y)}^2} - 2} \over 2}} \right| = 2(x + y)$$
B
$$\ln \left| {{{{{(x + y)}^2} - 2} \over 2}} \right| = 2(x - y)$$
C
$$\ln \left| {{{{{(x - y)}^2} + 2} \over 2}} \right| = 2(x + y)$$
D
$$\ln \left| {{{{{(x + y)}^2} + 2} \over 2}} \right| = 2(x + y)$$
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