1
BITSAT 2023
+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
$$y=\frac{\log |\cos x|}{1-x^2}+\frac{C}{1-x^2}$$
2
BITSAT 2022
+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
$$y = {\tan ^2}x + \sin x + c$$
3
BITSAT 2021
+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
$$\ln \left| {{{{{(x + y)}^2} + 2} \over 2}} \right| = 2(x + y)$$
4
BITSAT 2021
+3
-1

The solution of $${x^3}{{dy} \over {dx}} + 4{x^2}\tan y = {e^x}\sec y$$ satisfying y (1) = 0, is

A
$$\tan y = (x - 2){e^x}\log x$$
B
$$\sin y = {e^x}(x - 1){x^{ - 4}}$$
C
$$\tan y = (x - 1){e^x}{x^{ - 3}}$$
D
$$\sin y = {e^x}(x - 1){x^3}$$
EXAM MAP
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