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JEE Main Previous Year Questions (2026): Differential Equations
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Page 1
JEE Main Previous Year Questions
(2025): Differential Equations
Question1: Let ?? = ?? ( ?? ) be the solution of the differential equation
?? ?? ?? ?? +
????
?? ?? -?? =
?? ?? +?? ?? v
?? -?? ?? , -?? < ?? < ?? such that ?? ( ?? )= ?? . If ?? ? ?
?? /?? -?? /?? ?? ( ?? ) ?? ?? = ?? ?? - ?? then ?? ?? is equal
to ____ .
JEE Main 2025 (Online) 22nd January Evening Shift
Answer: 27
Solution:
I.F. e
-
1
2
?
2x
1-x
2
dx
= e
-
1
2
ln ( 1-x
2
)
= v1 - x
2
?? ×
v
1 - ?? 2
= ? ( ?? 6
+ 4?? ) ???? =
?? 7
7
+ 2?? 2
+ ??
Given ?? ( 0) = 0 ? ?? = 0
?? =
?? 7
7
+ 2?? 2
v1 - ?? 2
Now, 6?
-
1
2
1
2
?
?? 7
7
+2?? 2
v1-?? 2
???? = 6?
-
1
2
1
2
?
2?? 2
v1-?? 2
????
= 24?
0
1
2
?
x
2
v1 - x
2
dx
Put ?? = sin ??
???? = cos ??????
= 24?
0
?? 6
?
sin
2
?? cos ?? cos ??????
= 24?
0
?? 6
?(
1 - cos 2?? 2
)???? = 12[?? -
sin 2?? 2
]
0
?? 6
= 12(
?? 6
-
v3
4
)
= 2?? - 3v3
?? 2
= ( 3v3)
2
= 27
Question2: Let ?? be a differentiable function such that
?? ( ?? + ?? )
?? ?? ( ?? )- ?? ( ?? + ?? )
?? = ???? ?
?? ?? ?( ?? + ?? ) ?? ( ?? ) ???? , ?? = ?? . Then ?? ( ?? ) is equal to ____ .
JEE Main 2025 (Online) 24th January Morning Shift
Answer: 19
Page 2
JEE Main Previous Year Questions
(2025): Differential Equations
Question1: Let ?? = ?? ( ?? ) be the solution of the differential equation
?? ?? ?? ?? +
????
?? ?? -?? =
?? ?? +?? ?? v
?? -?? ?? , -?? < ?? < ?? such that ?? ( ?? )= ?? . If ?? ? ?
?? /?? -?? /?? ?? ( ?? ) ?? ?? = ?? ?? - ?? then ?? ?? is equal
to ____ .
JEE Main 2025 (Online) 22nd January Evening Shift
Answer: 27
Solution:
I.F. e
-
1
2
?
2x
1-x
2
dx
= e
-
1
2
ln ( 1-x
2
)
= v1 - x
2
?? ×
v
1 - ?? 2
= ? ( ?? 6
+ 4?? ) ???? =
?? 7
7
+ 2?? 2
+ ??
Given ?? ( 0) = 0 ? ?? = 0
?? =
?? 7
7
+ 2?? 2
v1 - ?? 2
Now, 6?
-
1
2
1
2
?
?? 7
7
+2?? 2
v1-?? 2
???? = 6?
-
1
2
1
2
?
2?? 2
v1-?? 2
????
= 24?
0
1
2
?
x
2
v1 - x
2
dx
Put ?? = sin ??
???? = cos ??????
= 24?
0
?? 6
?
sin
2
?? cos ?? cos ??????
= 24?
0
?? 6
?(
1 - cos 2?? 2
)???? = 12[?? -
sin 2?? 2
]
0
?? 6
= 12(
?? 6
-
v3
4
)
= 2?? - 3v3
?? 2
= ( 3v3)
2
= 27
Question2: Let ?? be a differentiable function such that
?? ( ?? + ?? )
?? ?? ( ?? )- ?? ( ?? + ?? )
?? = ???? ?
?? ?? ?( ?? + ?? ) ?? ( ?? ) ???? , ?? = ?? . Then ?? ( ?? ) is equal to ____ .
JEE Main 2025 (Online) 24th January Morning Shift
Answer: 19
Solution:
Differentiate both sides
4( ?? + 2) ?? ( ?? )+ 2( ?? + 2)
2
?? '
( ?? )- 6( ?? + 2) = 10( ?? + 2) ?? ( ?? )
2( ?? + 2)
2
?? '
( ?? )- 6( ?? + 2) ?? ( ?? ) = 6( ?? + 2)
( ?? + 2)
????
????
- 3?? = 3
?
????
????
= 3?
????
?? + 2
ln ( ?? + 1) = 3ln ( ?? + 2)+ ??
( ?? + 1) = ?? ( ?? + 2)
3
?? ( 0) =
3
2
?? ( 2) = 19
Question3: Let ?? = ?? ( ?? ) be the solution of the differential equation
???????? ?? ?? ?? ?? ?? = ?????? ?? ?? - ?? ?? ?????? ?? , ?? ? (?? ,
?? ?? ). If ?? (
?? ?? ) = ?? , then ?? '
(
?? ?? ) + ?? (
?? ?? ) is equal to
____ .
JEE Main 2025 (Online) 24th January Evening Shift
Answer: 1
Solution:
????
????
+ 2?? tan ?? = sin ??
I.F. = ?? 2? tan ?????? = sec
2
??
?? sec
2
?? = ?
sin ?? cos
2
?? ????
= ? tan ?? sec ??????
= sec ?? + ??
?? = -2
?? = cos ?? - 2cos
2
??
?? (
?? 4
) =
1
v2
- 1
?? '
= -sin ?? + 4cos
2
?? sin ??
?? '
(
?? 4
) = -
1
v2
+ 2
?? '
(
?? 4
) + ?? (
?? 4
) = 1
Question4: If ?? = ?? ( ?? ) is the solution of the differential equation, v?? - ?? ?? ?? ?? ?? ?? =
((?????? -?? (
?? ?? ))
?? - ?? ) ?????? -?? (
?? ?? ), -?? = ?? = ?? , ?? ( ?? )=
?? ?? -?? ?? , then ?? ?? ( ?? ) is equal to __ ?
Page 3
JEE Main Previous Year Questions
(2025): Differential Equations
Question1: Let ?? = ?? ( ?? ) be the solution of the differential equation
?? ?? ?? ?? +
????
?? ?? -?? =
?? ?? +?? ?? v
?? -?? ?? , -?? < ?? < ?? such that ?? ( ?? )= ?? . If ?? ? ?
?? /?? -?? /?? ?? ( ?? ) ?? ?? = ?? ?? - ?? then ?? ?? is equal
to ____ .
JEE Main 2025 (Online) 22nd January Evening Shift
Answer: 27
Solution:
I.F. e
-
1
2
?
2x
1-x
2
dx
= e
-
1
2
ln ( 1-x
2
)
= v1 - x
2
?? ×
v
1 - ?? 2
= ? ( ?? 6
+ 4?? ) ???? =
?? 7
7
+ 2?? 2
+ ??
Given ?? ( 0) = 0 ? ?? = 0
?? =
?? 7
7
+ 2?? 2
v1 - ?? 2
Now, 6?
-
1
2
1
2
?
?? 7
7
+2?? 2
v1-?? 2
???? = 6?
-
1
2
1
2
?
2?? 2
v1-?? 2
????
= 24?
0
1
2
?
x
2
v1 - x
2
dx
Put ?? = sin ??
???? = cos ??????
= 24?
0
?? 6
?
sin
2
?? cos ?? cos ??????
= 24?
0
?? 6
?(
1 - cos 2?? 2
)???? = 12[?? -
sin 2?? 2
]
0
?? 6
= 12(
?? 6
-
v3
4
)
= 2?? - 3v3
?? 2
= ( 3v3)
2
= 27
Question2: Let ?? be a differentiable function such that
?? ( ?? + ?? )
?? ?? ( ?? )- ?? ( ?? + ?? )
?? = ???? ?
?? ?? ?( ?? + ?? ) ?? ( ?? ) ???? , ?? = ?? . Then ?? ( ?? ) is equal to ____ .
JEE Main 2025 (Online) 24th January Morning Shift
Answer: 19
Solution:
Differentiate both sides
4( ?? + 2) ?? ( ?? )+ 2( ?? + 2)
2
?? '
( ?? )- 6( ?? + 2) = 10( ?? + 2) ?? ( ?? )
2( ?? + 2)
2
?? '
( ?? )- 6( ?? + 2) ?? ( ?? ) = 6( ?? + 2)
( ?? + 2)
????
????
- 3?? = 3
?
????
????
= 3?
????
?? + 2
ln ( ?? + 1) = 3ln ( ?? + 2)+ ??
( ?? + 1) = ?? ( ?? + 2)
3
?? ( 0) =
3
2
?? ( 2) = 19
Question3: Let ?? = ?? ( ?? ) be the solution of the differential equation
???????? ?? ?? ?? ?? ?? = ?????? ?? ?? - ?? ?? ?????? ?? , ?? ? (?? ,
?? ?? ). If ?? (
?? ?? ) = ?? , then ?? '
(
?? ?? ) + ?? (
?? ?? ) is equal to
____ .
JEE Main 2025 (Online) 24th January Evening Shift
Answer: 1
Solution:
????
????
+ 2?? tan ?? = sin ??
I.F. = ?? 2? tan ?????? = sec
2
??
?? sec
2
?? = ?
sin ?? cos
2
?? ????
= ? tan ?? sec ??????
= sec ?? + ??
?? = -2
?? = cos ?? - 2cos
2
??
?? (
?? 4
) =
1
v2
- 1
?? '
= -sin ?? + 4cos
2
?? sin ??
?? '
(
?? 4
) = -
1
v2
+ 2
?? '
(
?? 4
) + ?? (
?? 4
) = 1
Question4: If ?? = ?? ( ?? ) is the solution of the differential equation, v?? - ?? ?? ?? ?? ?? ?? =
((?????? -?? (
?? ?? ))
?? - ?? ) ?????? -?? (
?? ?? ), -?? = ?? = ?? , ?? ( ?? )=
?? ?? -?? ?? , then ?? ?? ( ?? ) is equal to __ ?
JEE Main 2025 (Online) 28th January Evening Shift
Answer: 4
Solution:
????
????
+
(sin
-1
?? 2
)
v4 - ?? 2
?? =
(sin
-3
?? 2
)
3
v4 - ?? 2
?? ?? (sin
-1
?? 2
)
2
2
= ?
(sin
-3
?? 2
)
3
4 - ?? 2
?? (sin
-1
?? 2
)
2
2
????
?? = (sin
-1
?? 2
)
2
- 2 + ?? · ?? -(sin
-1
?? 2
)
2
2
?? ( 2) =
?? 2
4
- 2 ? ?? = 0
?? ( 0) = -2
Question5: Let ?? = ?? ( ?? ) be the solution of the differential equation
?? ?? ?? ?? + ?? ?? ?????? ?? ?? = ?? ?????? ?? ?? + ???????? ?? · ?????? ?? ?? ???????? that ?? ( ?? ) =
?? ?? . Then ???? (?? (
?? ?? ) -
?? -?? ) is equal to ____
JEE Main 2025 (Online) 2nd April Evening Shift
Answer: 21
Solution:
????
????
+ 2?? sec
2
?? = 2sec
2
?? + 3tan ?? sec
2
??
I.F. = ?? ? 2sec
2
??????
I.F. = ?? 2tan ??
?? · ?? 2tan ?? = ? ?? 2tan ?? ( 2 + 3tan ?? ) sec
2
??????
Put tan ?? = ??
sec
2
?????? = ????
?? · ?? 2?? = ? ?? 2?? ( 2 + 3?? ) ????
?? · ?? 2?? ?
2?? 2?? 2
+ 3? ?? 2?? · ??????
?? · ?? 2?? = ?? 2?? + 3 [
?? ?? 2?? 2
- ?
?? 2?? 2
]
?? ?? 2?? = ?? 2?? + 3 [
?? ?? 2?? 2
-
?? 2?? 4
] + ??
?? ?? 2tan ?? = ?? 2tan ?? + 3 [
tan ?? ?? 2tan ?? 2
-
?? 2tan ?? 4
] + ??
?? ( 0) =
5
4
Page 4
JEE Main Previous Year Questions
(2025): Differential Equations
Question1: Let ?? = ?? ( ?? ) be the solution of the differential equation
?? ?? ?? ?? +
????
?? ?? -?? =
?? ?? +?? ?? v
?? -?? ?? , -?? < ?? < ?? such that ?? ( ?? )= ?? . If ?? ? ?
?? /?? -?? /?? ?? ( ?? ) ?? ?? = ?? ?? - ?? then ?? ?? is equal
to ____ .
JEE Main 2025 (Online) 22nd January Evening Shift
Answer: 27
Solution:
I.F. e
-
1
2
?
2x
1-x
2
dx
= e
-
1
2
ln ( 1-x
2
)
= v1 - x
2
?? ×
v
1 - ?? 2
= ? ( ?? 6
+ 4?? ) ???? =
?? 7
7
+ 2?? 2
+ ??
Given ?? ( 0) = 0 ? ?? = 0
?? =
?? 7
7
+ 2?? 2
v1 - ?? 2
Now, 6?
-
1
2
1
2
?
?? 7
7
+2?? 2
v1-?? 2
???? = 6?
-
1
2
1
2
?
2?? 2
v1-?? 2
????
= 24?
0
1
2
?
x
2
v1 - x
2
dx
Put ?? = sin ??
???? = cos ??????
= 24?
0
?? 6
?
sin
2
?? cos ?? cos ??????
= 24?
0
?? 6
?(
1 - cos 2?? 2
)???? = 12[?? -
sin 2?? 2
]
0
?? 6
= 12(
?? 6
-
v3
4
)
= 2?? - 3v3
?? 2
= ( 3v3)
2
= 27
Question2: Let ?? be a differentiable function such that
?? ( ?? + ?? )
?? ?? ( ?? )- ?? ( ?? + ?? )
?? = ???? ?
?? ?? ?( ?? + ?? ) ?? ( ?? ) ???? , ?? = ?? . Then ?? ( ?? ) is equal to ____ .
JEE Main 2025 (Online) 24th January Morning Shift
Answer: 19
Solution:
Differentiate both sides
4( ?? + 2) ?? ( ?? )+ 2( ?? + 2)
2
?? '
( ?? )- 6( ?? + 2) = 10( ?? + 2) ?? ( ?? )
2( ?? + 2)
2
?? '
( ?? )- 6( ?? + 2) ?? ( ?? ) = 6( ?? + 2)
( ?? + 2)
????
????
- 3?? = 3
?
????
????
= 3?
????
?? + 2
ln ( ?? + 1) = 3ln ( ?? + 2)+ ??
( ?? + 1) = ?? ( ?? + 2)
3
?? ( 0) =
3
2
?? ( 2) = 19
Question3: Let ?? = ?? ( ?? ) be the solution of the differential equation
???????? ?? ?? ?? ?? ?? = ?????? ?? ?? - ?? ?? ?????? ?? , ?? ? (?? ,
?? ?? ). If ?? (
?? ?? ) = ?? , then ?? '
(
?? ?? ) + ?? (
?? ?? ) is equal to
____ .
JEE Main 2025 (Online) 24th January Evening Shift
Answer: 1
Solution:
????
????
+ 2?? tan ?? = sin ??
I.F. = ?? 2? tan ?????? = sec
2
??
?? sec
2
?? = ?
sin ?? cos
2
?? ????
= ? tan ?? sec ??????
= sec ?? + ??
?? = -2
?? = cos ?? - 2cos
2
??
?? (
?? 4
) =
1
v2
- 1
?? '
= -sin ?? + 4cos
2
?? sin ??
?? '
(
?? 4
) = -
1
v2
+ 2
?? '
(
?? 4
) + ?? (
?? 4
) = 1
Question4: If ?? = ?? ( ?? ) is the solution of the differential equation, v?? - ?? ?? ?? ?? ?? ?? =
((?????? -?? (
?? ?? ))
?? - ?? ) ?????? -?? (
?? ?? ), -?? = ?? = ?? , ?? ( ?? )=
?? ?? -?? ?? , then ?? ?? ( ?? ) is equal to __ ?
JEE Main 2025 (Online) 28th January Evening Shift
Answer: 4
Solution:
????
????
+
(sin
-1
?? 2
)
v4 - ?? 2
?? =
(sin
-3
?? 2
)
3
v4 - ?? 2
?? ?? (sin
-1
?? 2
)
2
2
= ?
(sin
-3
?? 2
)
3
4 - ?? 2
?? (sin
-1
?? 2
)
2
2
????
?? = (sin
-1
?? 2
)
2
- 2 + ?? · ?? -(sin
-1
?? 2
)
2
2
?? ( 2) =
?? 2
4
- 2 ? ?? = 0
?? ( 0) = -2
Question5: Let ?? = ?? ( ?? ) be the solution of the differential equation
?? ?? ?? ?? + ?? ?? ?????? ?? ?? = ?? ?????? ?? ?? + ???????? ?? · ?????? ?? ?? ???????? that ?? ( ?? ) =
?? ?? . Then ???? (?? (
?? ?? ) -
?? -?? ) is equal to ____
JEE Main 2025 (Online) 2nd April Evening Shift
Answer: 21
Solution:
????
????
+ 2?? sec
2
?? = 2sec
2
?? + 3tan ?? sec
2
??
I.F. = ?? ? 2sec
2
??????
I.F. = ?? 2tan ??
?? · ?? 2tan ?? = ? ?? 2tan ?? ( 2 + 3tan ?? ) sec
2
??????
Put tan ?? = ??
sec
2
?????? = ????
?? · ?? 2?? = ? ?? 2?? ( 2 + 3?? ) ????
?? · ?? 2?? ?
2?? 2?? 2
+ 3? ?? 2?? · ??????
?? · ?? 2?? = ?? 2?? + 3 [
?? ?? 2?? 2
- ?
?? 2?? 2
]
?? ?? 2?? = ?? 2?? + 3 [
?? ?? 2?? 2
-
?? 2?? 4
] + ??
?? ?? 2tan ?? = ?? 2tan ?? + 3 [
tan ?? ?? 2tan ?? 2
-
?? 2tan ?? 4
] + ??
?? ( 0) =
5
4
5
4
= 1 -
3
4
+ ??
5
4
-
1
4
= ??
1 = ??
?? = 1 + 3 (
tan ?? 2
-
1
4
) + 1 · ?? -2tan ??
?? (
?? 4
) = 1 + 3 (
1
2
-
1
4
) +
1
?? 2
?? (
?? 4
) =
7
4
+
1
?? 2
12(?? (
?? 4
) -
1
?? 2
) = 12(
7
4
+
1
?? 2
-
1
?? 2
) = 21
Question6: Let ?? : ?? ? ?? be a twice differentiable function such that ?? ( ?? + ?? ) =
?? ( ?? ) ?? ( ?? ) for all ?? , ?? ? ?? . If ?? '
( ?? ) = ???? and ?? satisfies ?? ''
( ?? )- ???? ?? '
( ?? )- ?? ( ?? )=
?? , ?? > ?? , then the area of the region ?? = {( ?? , ?? ) | ?? = ?? = ?? ( ???? ) , ?? = ?? = ?? } is :
JEE Main 2025 (Online) 22nd January Morning Shift
Options:
A. e
2
- 1
B. ?? 4
+ 1
C. e
2
+ 1
D. ?? 4
- 1
Answer: A
Solution:
f( x + y) = f( x ) f( y)
? f( x ) = e
?? x
f
'
( 0) = 4a
? f
'
( x ) = ?? e
?? x
? ?? = 4a
So, ?? ( ?? ) = ?? 4????
f
''
( x )- 3af
'
( x )- f( x ) = 0
? ?? 2
- 3a ?? - 1 = 0
? 16a
2
- 12a
2
- 1 = 0 ? 4a
2
= 1 ? a =
1
2
Page 5
JEE Main Previous Year Questions
(2025): Differential Equations
Question1: Let ?? = ?? ( ?? ) be the solution of the differential equation
?? ?? ?? ?? +
????
?? ?? -?? =
?? ?? +?? ?? v
?? -?? ?? , -?? < ?? < ?? such that ?? ( ?? )= ?? . If ?? ? ?
?? /?? -?? /?? ?? ( ?? ) ?? ?? = ?? ?? - ?? then ?? ?? is equal
to ____ .
JEE Main 2025 (Online) 22nd January Evening Shift
Answer: 27
Solution:
I.F. e
-
1
2
?
2x
1-x
2
dx
= e
-
1
2
ln ( 1-x
2
)
= v1 - x
2
?? ×
v
1 - ?? 2
= ? ( ?? 6
+ 4?? ) ???? =
?? 7
7
+ 2?? 2
+ ??
Given ?? ( 0) = 0 ? ?? = 0
?? =
?? 7
7
+ 2?? 2
v1 - ?? 2
Now, 6?
-
1
2
1
2
?
?? 7
7
+2?? 2
v1-?? 2
???? = 6?
-
1
2
1
2
?
2?? 2
v1-?? 2
????
= 24?
0
1
2
?
x
2
v1 - x
2
dx
Put ?? = sin ??
???? = cos ??????
= 24?
0
?? 6
?
sin
2
?? cos ?? cos ??????
= 24?
0
?? 6
?(
1 - cos 2?? 2
)???? = 12[?? -
sin 2?? 2
]
0
?? 6
= 12(
?? 6
-
v3
4
)
= 2?? - 3v3
?? 2
= ( 3v3)
2
= 27
Question2: Let ?? be a differentiable function such that
?? ( ?? + ?? )
?? ?? ( ?? )- ?? ( ?? + ?? )
?? = ???? ?
?? ?? ?( ?? + ?? ) ?? ( ?? ) ???? , ?? = ?? . Then ?? ( ?? ) is equal to ____ .
JEE Main 2025 (Online) 24th January Morning Shift
Answer: 19
Solution:
Differentiate both sides
4( ?? + 2) ?? ( ?? )+ 2( ?? + 2)
2
?? '
( ?? )- 6( ?? + 2) = 10( ?? + 2) ?? ( ?? )
2( ?? + 2)
2
?? '
( ?? )- 6( ?? + 2) ?? ( ?? ) = 6( ?? + 2)
( ?? + 2)
????
????
- 3?? = 3
?
????
????
= 3?
????
?? + 2
ln ( ?? + 1) = 3ln ( ?? + 2)+ ??
( ?? + 1) = ?? ( ?? + 2)
3
?? ( 0) =
3
2
?? ( 2) = 19
Question3: Let ?? = ?? ( ?? ) be the solution of the differential equation
???????? ?? ?? ?? ?? ?? = ?????? ?? ?? - ?? ?? ?????? ?? , ?? ? (?? ,
?? ?? ). If ?? (
?? ?? ) = ?? , then ?? '
(
?? ?? ) + ?? (
?? ?? ) is equal to
____ .
JEE Main 2025 (Online) 24th January Evening Shift
Answer: 1
Solution:
????
????
+ 2?? tan ?? = sin ??
I.F. = ?? 2? tan ?????? = sec
2
??
?? sec
2
?? = ?
sin ?? cos
2
?? ????
= ? tan ?? sec ??????
= sec ?? + ??
?? = -2
?? = cos ?? - 2cos
2
??
?? (
?? 4
) =
1
v2
- 1
?? '
= -sin ?? + 4cos
2
?? sin ??
?? '
(
?? 4
) = -
1
v2
+ 2
?? '
(
?? 4
) + ?? (
?? 4
) = 1
Question4: If ?? = ?? ( ?? ) is the solution of the differential equation, v?? - ?? ?? ?? ?? ?? ?? =
((?????? -?? (
?? ?? ))
?? - ?? ) ?????? -?? (
?? ?? ), -?? = ?? = ?? , ?? ( ?? )=
?? ?? -?? ?? , then ?? ?? ( ?? ) is equal to __ ?
JEE Main 2025 (Online) 28th January Evening Shift
Answer: 4
Solution:
????
????
+
(sin
-1
?? 2
)
v4 - ?? 2
?? =
(sin
-3
?? 2
)
3
v4 - ?? 2
?? ?? (sin
-1
?? 2
)
2
2
= ?
(sin
-3
?? 2
)
3
4 - ?? 2
?? (sin
-1
?? 2
)
2
2
????
?? = (sin
-1
?? 2
)
2
- 2 + ?? · ?? -(sin
-1
?? 2
)
2
2
?? ( 2) =
?? 2
4
- 2 ? ?? = 0
?? ( 0) = -2
Question5: Let ?? = ?? ( ?? ) be the solution of the differential equation
?? ?? ?? ?? + ?? ?? ?????? ?? ?? = ?? ?????? ?? ?? + ???????? ?? · ?????? ?? ?? ???????? that ?? ( ?? ) =
?? ?? . Then ???? (?? (
?? ?? ) -
?? -?? ) is equal to ____
JEE Main 2025 (Online) 2nd April Evening Shift
Answer: 21
Solution:
????
????
+ 2?? sec
2
?? = 2sec
2
?? + 3tan ?? sec
2
??
I.F. = ?? ? 2sec
2
??????
I.F. = ?? 2tan ??
?? · ?? 2tan ?? = ? ?? 2tan ?? ( 2 + 3tan ?? ) sec
2
??????
Put tan ?? = ??
sec
2
?????? = ????
?? · ?? 2?? = ? ?? 2?? ( 2 + 3?? ) ????
?? · ?? 2?? ?
2?? 2?? 2
+ 3? ?? 2?? · ??????
?? · ?? 2?? = ?? 2?? + 3 [
?? ?? 2?? 2
- ?
?? 2?? 2
]
?? ?? 2?? = ?? 2?? + 3 [
?? ?? 2?? 2
-
?? 2?? 4
] + ??
?? ?? 2tan ?? = ?? 2tan ?? + 3 [
tan ?? ?? 2tan ?? 2
-
?? 2tan ?? 4
] + ??
?? ( 0) =
5
4
5
4
= 1 -
3
4
+ ??
5
4
-
1
4
= ??
1 = ??
?? = 1 + 3 (
tan ?? 2
-
1
4
) + 1 · ?? -2tan ??
?? (
?? 4
) = 1 + 3 (
1
2
-
1
4
) +
1
?? 2
?? (
?? 4
) =
7
4
+
1
?? 2
12(?? (
?? 4
) -
1
?? 2
) = 12(
7
4
+
1
?? 2
-
1
?? 2
) = 21
Question6: Let ?? : ?? ? ?? be a twice differentiable function such that ?? ( ?? + ?? ) =
?? ( ?? ) ?? ( ?? ) for all ?? , ?? ? ?? . If ?? '
( ?? ) = ???? and ?? satisfies ?? ''
( ?? )- ???? ?? '
( ?? )- ?? ( ?? )=
?? , ?? > ?? , then the area of the region ?? = {( ?? , ?? ) | ?? = ?? = ?? ( ???? ) , ?? = ?? = ?? } is :
JEE Main 2025 (Online) 22nd January Morning Shift
Options:
A. e
2
- 1
B. ?? 4
+ 1
C. e
2
+ 1
D. ?? 4
- 1
Answer: A
Solution:
f( x + y) = f( x ) f( y)
? f( x ) = e
?? x
f
'
( 0) = 4a
? f
'
( x ) = ?? e
?? x
? ?? = 4a
So, ?? ( ?? ) = ?? 4????
f
''
( x )- 3af
'
( x )- f( x ) = 0
? ?? 2
- 3a ?? - 1 = 0
? 16a
2
- 12a
2
- 1 = 0 ? 4a
2
= 1 ? a =
1
2
F( x) = e
2x
Area = ?
0
2
?e
x
dx= e
2
- 1
Question7: Let ?? ( ?? ) be a real differentiable function such that ?? ( ?? )= ?? and ?? ( ?? +
?? ) = ?? ( ?? ) ?? '
( ?? )+ ?? '
( ?? ) ?? ( ?? ) for all ?? , ?? ? ?? . Then ?
?? =?? ??????
??????? ?? ?? ( ?? ) is equal to:
JEE Main 2025 (Online) 22nd January Morning Shift
Options:
A. 2406
B. 5220
C. 2525
D. 2384
Answer: C
Solution:
?? ( ?? + ?? ) = ?? ( ?? ) ?? '
( ?? )+ ?? '
( ?? ) ?? ( ?? )
Put = ?? = ?? = 0
?? ( 0) = ?? ( 0) ?? '
( 0)+ ?? '
( 0) ?? ( 0)
?? '
( 0) =
1
2
Put ?? = 0
?? ( ?? ) = ?? ( ?? ) ?? '
( 0)+ ?? '
( ?? ) ?? ( 0)
?? ( ?? ) =
1
2
?? ( ?? )+ ?? '
( ?? )
?? '
( ?? ) =
?? ( ?? )
2
Read MoreFAQs on JEE Main Previous Year Questions (2026): Differential Equations
| 1. What are the key concepts of differential equations that JEE aspirants should focus on? |  |
Ans. Aspirants should focus on the following key concepts: types of differential equations (ordinary and partial), order and degree, methods of solving first-order differential equations (separation of variables, integrating factor), higher-order linear differential equations, and applications of differential equations in real-life problems. Understanding initial and boundary value problems is also crucial.
| 2. How can one effectively prepare for differential equations in JEE? | |
Ans. Effective preparation includes a thorough understanding of the theory and concepts, practicing a variety of problems from previous years and sample papers, focusing on derivations and applications, and using online resources or coaching for additional guidance. Regular revision and solving mock tests can also greatly enhance performance.
| 3. What are the common types of questions related to differential equations asked in JEE? | |
Ans. Common questions include finding general and particular solutions to differential equations, solving initial value problems, applying Laplace transforms, and interpreting real-world scenarios modeled by differential equations. Questions may also involve verifying solutions and applying specific methods for solving different forms of equations.
| 4. Are there any important theorems or formulas related to differential equations that JEE students should memorize? | |
Ans. Yes, students should memorize key theorems and formulas such as the existence and uniqueness theorem for solutions of differential equations, the linearity principle for linear differential equations, and the method of undetermined coefficients. Additionally, understanding the Wronskian for linear independence of solutions is also important.
| 5. How important is the application of differential equations in JEE and in real life? | |
Ans. The application of differential equations is quite important in JEE as it demonstrates the ability to model and solve real-life problems in physics, engineering, and other fields. Understanding how to apply these concepts helps students in various topics, including dynamics, heat transfer, and population modeling, thus enhancing both their analytical skills and practical knowledge.
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