Question 1: Palette 1/9

BITSAT 2025 MCQ

The solution of the differential equation $\frac{d y}{d x}+x \sin 2 y=x^3 \cos ^2 y$ is

A

$\tan y=\left(x^2-1\right) e^{x^2}+C$

B

$e^{x^2}=\frac{1}{2}\left(x^2-1\right) \tan y e^{x^2}+C$

C

$\tan y\left(x^2-1\right)=e^{x^2}+C$

D

$e^{x^2} \tan y=\frac{1}{2}\left(x^2-1\right) e^{x^2}+C$

Question 2: Palette 2/9

BITSAT 2024 MCQ

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 $

Question 3: Palette 3/9

BITSAT 2023 MCQ

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

$y=\frac{\log |\cos x|}{1-x^2}+\frac{C}{1-x^2}$

Question 4: Palette 4/9

BITSAT 2022 MCQ

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

$y = {\tan ^2}x + \sin x + c$

Question 5: Palette 5/9

BITSAT 2021 MCQ

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

$\ln \left| {{{{{(x + y)}^2} + 2} \over 2}} \right| = 2(x + y)$

Question 6: Palette 6/9

BITSAT 2021 MCQ

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

$\sin y = {e^x}(x - 1){x^3}$

Question 7: Palette 7/9

BITSAT 2020 MCQ

The solution of the equation ${{dy} \over {dx}} + {1 \over x}\tan y = {1 \over {{x^2}}}\tan y\sin y$ is

A

$2y = \sin y(1 - 2c{x^2})$

B

$2x = \cot y(1 + 2c{x^2})$

C

$2x = \sin y(1 - 2c{x^2})$

D

$2x\sin y = 1 - 2c{x^2}$

Question 8: Palette 8/9

BITSAT 2020 MCQ

The solution of differential equation $(x{y^5} + 2y)dx - xdy = 0$, is

A

$9{x^8} + 4{x^9}{y^4} = 9{y^4}C$

B

$9{x^8} - 4{x^9}{y^4} - 9{y^4}C = 0$

C

${x^8}(9 + 4{y^4}) = 10{y^4}C$

D

None of these

Question 9: Palette 9/9

BITSAT 2020 MCQ

A curve passes through (2, 0) and the slope of the tangent at P(x, y) is equal to ${{{{(x + 1)}^2} + y - 3} \over {x + 1}}$ then the equation of the curve is

A

y = x² $-$ 2x

B

y = x³ $-$ 8

C

y² = x² + 2x

D

y² = 5x² $-$ 6