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Previous year question hub

Differential Equations - Calculus - Mathematics Previous Year Questions

Practice Differential Equations - Calculus - Mathematics previous year questions organised from real papers, with year-wise coverage and clear topic navigation.

6Papers
6Years
9Questions
1Topics

Differential Equations question pattern

Every graph below is calculated only from this selection.

Questions by year

Year-wise coverage for Differential Equations. Each bar uses a separate theme-derived color.

Difficulty distribution

How the classified questions are distributed by difficulty.

Not classified 9 100%

Question type distribution

MCQ, numerical, multiple-select and other formats found in these papers.

Multiple Choices 9 100%

Subject weightage

Top subjects by unique question coverage.

Mathematics
9 Qs

Most asked topics

Top topics across the included previous year papers.

Calculus
9 Qs

Subtopic coverage

Top subtopics inside this exact selection.

Differential Equations
9 Qs

Paper coverage

Question coverage for the most populated papers. Every active PYP paper remains listed below.

BITSAT 2025
1 Qs
BITSAT 2024
1 Qs
BITSAT 2023
1 Qs
BITSAT 2022
1 Qs
BITSAT 2021
2 Qs
BITSAT 2020
3 Qs

Included previous year papers

Newest papers appear first. Sort by year, question coverage or name.

PaperYear / sessionQuestions in this viewOpen
BITSAT 202520251View paper
BITSAT 202420241View paper
BITSAT 202320231View paper
BITSAT 202220221View paper
BITSAT 202120212View paper
BITSAT 202020203View paper

All Differential Equations previous year questions

Practice every matching question in batches of 20, with every available option.

1
2020 · Mathematics · Calculus · Differential Equations
BITSAT 2020

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
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2
2020 · Mathematics · Calculus · Differential Equations
BITSAT 2020

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
\(2x\sin y = 1 - 2c{x^2}\)
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3
2020 · Mathematics · Calculus · Differential Equations
BITSAT 2020

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
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4
2021 · Mathematics · Calculus · Differential Equations
BITSAT 2021

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
\(\sin y = {e^x}(x - 1){x^3}\)
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5
2021 · Mathematics · Calculus · Differential Equations
BITSAT 2021

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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6
2022 · Mathematics · Calculus · Differential Equations
BITSAT 2022

\(\left( {{{dy} \over {dx}}} \right)\tan x = y{\sec ^2}x + \sin x\), 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
\(y = {\tan ^2}x + \sin x + c\)
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7
2023 · Mathematics · Calculus · Differential Equations
BITSAT 2023

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
\(y=\frac{\log |\cos x|}{1-x^2}+\frac{C}{1-x^2}\)
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8
2024 · Mathematics · Calculus · Differential Equations
BITSAT 2024

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 $
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9
2025 · Mathematics · Calculus · Differential Equations
BITSAT 2025

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

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

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