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Test Index
JEE Main 26 June 2022 Shift 2 Solved Paper
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© examsnet.com
Question : 10
Total: 90
If the solution of the differential equation
d
y
d
x
+
e
x
(
x
2
−
2
)
y
=
(
x
2
−
2
x
)
(
x
2
−
2
)
e
2
x
satisfies
y
(
0
)
=
0
, then the value of
y
(
2
)
is__
[26-Jun-2022-Shift-2]
−
1
1
0
e
Validate
Solution:
👈: Video Solution
∵
d
y
d
x
+
e
x
(
x
2
−
2
)
y
=
(
x
2
−
2
x
)
(
x
2
−
2
)
e
2
x
Here,
I
.
F
.
=
e
∫
e
x
(
x
2
−
2
)
d
x
=
e
(
x
2
−
2
x
)
e
x
∴
Solution of the differential equation is
y
⋅
e
(
x
2
−
2
x
)
e
x
=
∫
(
x
2
−
2
x
)
(
x
2
−
2
)
e
2
x
⋅
e
(
x
2
−
2
x
)
e
x
d
x
=
∫
(
x
2
−
2
x
)
e
x
⋅
(
x
2
−
2
)
e
x
⋅
e
(
x
2
−
2
x
)
e
x
d
x
Let
(
x
2
−
2
x
)
e
x
=
t
∴
(
x
2
−
2
)
e
x
d
x
=
d
t
y
⋅
e
(
x
2
−
2
x
)
e
x
=
∫
t
⋅
e
t
d
t
y
⋅
e
(
x
2
−
2
x
)
e
x
=
(
x
2
−
2
x
−
1
)
e
(
x
2
−
2
x
)
e
x
+
c
∴
y
(
0
)
=
0
∴
c
=
1
∴
y
=
(
x
2
−
2
x
−
1
)
+
e
(
2
x
−
x
2
)
e
x
∴
y
(
2
)
=
−
1
+
1
=
0
© examsnet.com
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